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2.2 Speed and Velocity

Section learning objectives.

By the end of this section, you will be able to do the following:

• Calculate the average speed of an object
• Relate displacement and average velocity

Teacher Support

The learning objectives in this section will help your students master the following standards:

• (B) describe and analyze motion in one dimension using equations with the concepts of distance, displacement, speed, average velocity, instantaneous velocity, and acceleration.

In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Position and Speed of an Object, as well as the following standards:

Section Key Terms

In this section, students will apply what they have learned about distance and displacement to the concepts of speed and velocity.

[BL] [OL] Before students read the section, ask them to give examples of ways they have heard the word speed used. Then ask them if they have heard the word velocity used. Explain that these words are often used interchangeably in everyday life, but their scientific definitions are different. Tell students that they will learn about these differences as they read the section.

[AL] Explain to students that velocity, like displacement, is a vector quantity. Ask them to speculate about ways that speed is different from velocity. After they share their ideas, follow up with questions that deepen their thought process, such as: Why do you think that? What is an example? How might apply these terms to motion that you see every day?

There is more to motion than distance and displacement. Questions such as, “How long does a foot race take?” and “What was the runner’s speed?” cannot be answered without an understanding of other concepts. In this section we will look at time , speed, and velocity to expand our understanding of motion.

A description of how fast or slow an object moves is its speed. Speed is the rate at which an object changes its location. Like distance, speed is a scalar because it has a magnitude but not a direction. Because speed is a rate, it depends on the time interval of motion. You can calculate the elapsed time or the change in time, Δ t Δ t , of motion as the difference between the ending time and the beginning time

The SI unit of time is the second (s), and the SI unit of speed is meters per second (m/s), but sometimes kilometers per hour (km/h), miles per hour (mph) or other units of speed are used.

When you describe an object's speed, you often describe the average over a time period. Average speed , v avg , is the distance traveled divided by the time during which the motion occurs.

You can, of course, rearrange the equation to solve for either distance or time

Suppose, for example, a car travels 150 kilometers in 3.2 hours. Its average speed for the trip is

A car's speed would likely increase and decrease many times over a 3.2 hour trip. Its speed at a specific instant in time, however, is its instantaneous speed . A car's speedometer describes its instantaneous speed.

[OL] [AL] Caution students that average speed is not always the average of an object's initial and final speeds. For example, suppose a car travels a distance of 100 km. The first 50 km it travels 30 km/h and the second 50 km it travels at 60 km/h. Its average speed would be distance /(time interval) = (100 km)/[(50 km)/(30 km/h) + (50 km)/(60 km/h)] = 40 km/h. If the car had spent equal times at 30 km and 60 km rather than equal distances at these speeds, its average speed would have been 45 km/h.

[BL] [OL] Caution students that the terms speed, average speed, and instantaneous speed are all often referred to simply as speed in everyday language. Emphasize the importance in science to use correct terminology to avoid confusion and to properly communicate ideas.

Worked Example

Calculating average speed.

A marble rolls 5.2 m in 1.8 s. What was the marble's average speed?

We know the distance the marble travels, 5.2 m, and the time interval, 1.8 s. We can use these values in the average speed equation.

Average speed is a scalar, so we do not include direction in the answer. We can check the reasonableness of the answer by estimating: 5 meters divided by 2 seconds is 2.5 m/s. Since 2.5 m/s is close to 2.9 m/s, the answer is reasonable. This is about the speed of a brisk walk, so it also makes sense.

Practice Problems

A pitcher throws a baseball from the pitcher’s mound to home plate in 0.46 s. The distance is 18.4 m. What was the average speed of the baseball?

The vector version of speed is velocity. Velocity describes the speed and direction of an object. As with speed, it is useful to describe either the average velocity over a time period or the velocity at a specific moment. Average velocity is displacement divided by the time over which the displacement occurs.

Velocity, like speed, has SI units of meters per second (m/s), but because it is a vector, you must also include a direction. Furthermore, the variable v for velocity is bold because it is a vector, which is in contrast to the variable v for speed which is italicized because it is a scalar quantity.

Tips For Success

It is important to keep in mind that the average speed is not the same thing as the average velocity without its direction. Like we saw with displacement and distance in the last section, changes in direction over a time interval have a bigger effect on speed and velocity.

Suppose a passenger moved toward the back of a plane with an average velocity of –4 m/s. We cannot tell from the average velocity whether the passenger stopped momentarily or backed up before he got to the back of the plane. To get more details, we must consider smaller segments of the trip over smaller time intervals such as those shown in Figure 2.9 . If you consider infinitesimally small intervals, you can define instantaneous velocity , which is the velocity at a specific instant in time. Instantaneous velocity and average velocity are the same if the velocity is constant.

Earlier, you have read that distance traveled can be different than the magnitude of displacement. In the same way, speed can be different than the magnitude of velocity. For example, you drive to a store and return home in half an hour. If your car’s odometer shows the total distance traveled was 6 km, then your average speed was 12 km/h. Your average velocity, however, was zero because your displacement for the round trip is zero.

Watch Physics

Calculating average velocity or speed.

This video reviews vectors and scalars and describes how to calculate average velocity and average speed when you know displacement and change in time. The video also reviews how to convert km/h to m/s.

• A scalar quantity is fully described by its magnitude, while a vector needs both magnitude and direction to fully describe it. Displacement is an example of a scalar quantity and time is an example of a vector quantity.
• A scalar quantity is fully described by its magnitude, while a vector needs both magnitude and direction to fully describe it. Time is an example of a scalar quantity and displacement is an example of a vector quantity.
• A scalar quantity is fully described by its magnitude and direction, while a vector needs only magnitude to fully describe it. Displacement is an example of a scalar quantity and time is an example of a vector quantity.
• A scalar quantity is fully described by its magnitude and direction, while a vector needs only magnitude to fully describe it. Time is an example of a scalar quantity and displacement is an example of a vector quantity.

This video does a good job of reinforcing the difference between vectors and scalars. The student is introduced to the idea of using ‘s’ to denote displacement, which you may or may not wish to encourage. Before students watch the video, point out that the instructor uses s → s → for displacement instead of d, as used in this text. Explain the use of small arrows over variables is a common way to denote vectors in higher-level physics courses. Caution students that the customary abbreviations for hour and seconds are not used in this video. Remind students that in their own work they should use the abbreviations h for hour and s for seconds.

Calculating Average Velocity

A student has a displacement of 304 m north in 180 s. What was the student's average velocity?

We know that the displacement is 304 m north and the time is 180 s. We can use the formula for average velocity to solve the problem.

Since average velocity is a vector quantity, you must include direction as well as magnitude in the answer. Notice, however, that the direction can be omitted until the end to avoid cluttering the problem. Pay attention to the significant figures in the problem. The distance 304 m has three significant figures, but the time interval 180 s has only two, so the quotient should have only two significant figures.

Note the way scalars and vectors are represented. In this book d represents distance and displacement. Similarly, v represents speed, and v represents velocity. A variable that is not bold indicates a scalar quantity, and a bold variable indicates a vector quantity. Vectors are sometimes represented by small arrows above the variable.

Use this problem to emphasize the importance of using the correct number of significant figures in calculations. Some students have a tendency to include many digits in their final calculations. They incorrectly believe they are improving the accuracy of their answer by writing many of the digits shown on the calculator. Point out that doing this introduces errors into the calculations. In more complicated calculations, these errors can propagate and cause the final answer to be wrong. Instead, remind students to always carry one or two extra digits in intermediate calculations and to round the final answer to the correct number of significant figures.

Solving for Displacement when Average Velocity and Time are Known

Layla jogs with an average velocity of 2.4 m/s east. What is her displacement after 46 seconds?

We know that Layla's average velocity is 2.4 m/s east, and the time interval is 46 seconds. We can rearrange the average velocity formula to solve for the displacement.

The answer is about 110 m east, which is a reasonable displacement for slightly less than a minute of jogging. A calculator shows the answer as 110.4 m. We chose to write the answer using scientific notation because we wanted to make it clear that we only used two significant figures.

Dimensional analysis is a good way to determine whether you solved a problem correctly. Write the calculation using only units to be sure they match on opposite sides of the equal mark. In the worked example, you have m = (m/s)(s). Since seconds is in the denominator for the average velocity and in the numerator for the time, the unit cancels out leaving only m and, of course, m = m.

Solving for Time when Displacement and Average Velocity are Known

Phillip walks along a straight path from his house to his school. How long will it take him to get to school if he walks 428 m west with an average velocity of 1.7 m/s west?

We know that Phillip's displacement is 428 m west, and his average velocity is 1.7 m/s west. We can calculate the time required for the trip by rearranging the average velocity equation.

Here again we had to use scientific notation because the answer could only have two significant figures. Since time is a scalar, the answer includes only a magnitude and not a direction.

• 4 km/h north
• 4 km/h south
• 64 km/h north
• 64 km/h south

A bird flies with an average velocity of 7.5 m/s east from one branch to another in 2.4 s. It then pauses before flying with an average velocity of 6.8 m/s east for 3.5 s to another branch. What is the bird’s total displacement from its starting point?

Virtual Physics

The walking man.

In this simulation you will put your cursor on the man and move him first in one direction and then in the opposite direction. Keep the Introduction tab active. You can use the Charts tab after you learn about graphing motion later in this chapter. Carefully watch the sign of the numbers in the position and velocity boxes. Ignore the acceleration box for now. See if you can make the man’s position positive while the velocity is negative. Then see if you can do the opposite.

Grasp Check

Which situation correctly describes when the moving man’s position was negative but his velocity was positive?

• Man moving toward 0 from left of 0
• Man moving toward 0 from right of 0
• Man moving away from 0 from left of 0
• Man moving away from 0 from right of 0

This is a powerful interactive animation, and it can be used for many lessons. At this point it can be used to show that displacement can be either positive or negative. It can also show that when displacement is negative, velocity can be either positive or negative. Later it can be used to show that velocity and acceleration can have different signs. It is strongly suggested that you keep students on the Introduction tab. The Charts tab can be used after students learn about graphing motion later in this chapter.

Check Your Understanding

• Yes, because average velocity depends on the net or total displacement.
• Yes, because average velocity depends on the total distance traveled.
• No, because the velocities of both runners must remain exactly the same throughout the journey.
• No, because the instantaneous velocities of the runners must remain the same at the midpoint but can vary at other points.

If you divide the total distance traveled on a car trip (as determined by the odometer) by the time for the trip, are you calculating the average speed or the magnitude of the average velocity, and under what circumstances are these two quantities the same?

• Average speed. Both are the same when the car is traveling at a constant speed and changing direction.
• Average speed. Both are the same when the speed is constant and the car does not change its direction.
• Magnitude of average velocity. Both are same when the car is traveling at a constant speed.
• Magnitude of average velocity. Both are same when the car does not change its direction.
• Yes, if net displacement is negative.
• Yes, if the object’s direction changes during motion.
• No, because average velocity describes only the magnitude and not the direction of motion.
• No, because average velocity only describes the magnitude in the positive direction of motion.

Use the Check Your Understanding questions to assess students’ achievement of the sections learning objectives. If students are struggling with a specific objective, the Check Your Understanding will help identify which and direct students to the relevant content. Assessment items in TUTOR will allow you to reassess.

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Physics Problems with Solutions

Velocity and speed: solutions to problems.

Solutions to the problems on velocity and speed of moving objects. More tutorials can be found in this website.

A man walks 7 km in 2 hours and 2 km in 1 hour in the same direction. a) What is the man's average speed for the whole journey? b) What is the man's average velocity for the whole journey? Solution to Problem 1: a)

You start walking from a point on a circular field of radius 0.5 km and 1 hour later you are at the same point. a) What is your average speed for the whole journey? b) What is your average velocity for the whole journey? Solution to Problem 3: a) If you walk around a circular field and come back to the same point, you have covered a distance equal to the circumference of the circle.

John drove South 120 km at 60 km/h and then East 150 km at 50 km/h. Determine a) the average speed for the whole journey? b) the magnitude of the average velocity for the whole journey? Solution to Problem 4: a)

If I can walk at an average speed of 5 km/h, how many miles I can walk in two hours? Solution to Problem 5: distance = (average speed) * (time) = 5 km/h * 2 hours = 10 km using the rate of conversion 0.62 miles per km, the distance in miles is given by distance = 10 km * 0.62 miles/km = 6.2 miles

A train travels along a straight line at a constant speed of 60 mi/h for a distance d and then another distance equal to 2d in the same direction at a constant speed of 80 mi/h. a)What is the average speed of the train for the whole journey?

A car travels 22 km south, 12 km west, and 14 km north in half an hour. a) What is the average speed of the car? b) What is the final displacement of the car? c) What is the average velocity of the car?

Solution to Problem 7: a)

Solution to Problem 8: a)

More References and links

• Velocity and Speed: Tutorials with Examples
• Velocity and Speed: Problems with Solutions
• Acceleration: Tutorials with Examples
• Uniform Acceleration Motion: Problems with Solutions
• Uniform Acceleration Motion: Equations with Explanations

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Velocity is the rate at which an object changes position with time. An object is displaced when it changes its position. The amount of displacement over the time in which the displacement occurred gives the velocity. It is a vector quantity that has both magnitude and direction.

The importance of velocity is that it can give you an estimated time to go from one point to another. Suppose you are traveling from place A to place B. Velocity tells you how long it will take to arrive at your destination.

How to Calculate Velocity

Velocity can be calculated by measuring the object’s displacement over the time taken to displace it.

In vectorial notation, velocity is given by

The SI unit of velocity is Newtons per second of N/s. The cgs unit is ergs per second or ergs/s. Other units include miler per hour or mph, kilometers per hour or kph, and feet per second or fps.

• A car traveling on the highway at 60 mph toward the West
• A biker riding on the road at 35 mph toward the southwest
• A bicyclist heading northwards at 14 mph
• An airplane flying eastwards at 300 mph
• A rocket launched into the sky in the south-east direction
• A train traveling in the south direction at 80 kph
• A runner runs 420 m in 70 seconds

Average and Instantaneous Velocity

An object can move from one position to another in several steps. The average velocity is simply the change in position over the time interval within which change takes place.

v avg is the average velocity

x f is the final velocity

x i is the initial velocity

t f is the final time

t i is the initial time

If the starting time is zero, then t i = 0.

The instantaneous (v ins. ) velocity is the velocity at a particular instant of time. It can be obtained by taking the limit of the above expression in the limit Δt → 0.

Speed vs. Velocity

Both speed and velocity measure rate. Both of them depict how fast an object is moving. However, there is an essential difference. While speed tells us the magnitude of the rate at which the object moves, it does not say anything about the direction. On the other hand, velocity considers the direction of motion.

Let us take an example to illustrate this. When we say a car is moving on the road at 35 mph, we specify the magnitude but not the direction. However, when we say a car is moving on the road eastwards at 35 mph, we specify both the magnitude and direction. The first sentence refers to the speed, and the second refers to the velocity. Therefore, speed is a scalar quantity, and velocity is a vector quantity.

The formula for speed is the distance traveled over the time taken to travel that distance.

Velocity vs. Time Graph

The following graph shows how velocity changes with time. At some parts of the graph, the velocity increases linearly with time. The quotient of velocity and time is called acceleration . When the velocity decreases with time, the phenomenon is known as deceleration. Regions of constant velocity are also indicated on the graph.

Solved Problems

Problem 1. Suppose a runner moves along the x-direction over some time.During a 5.00 s interval, the runner’s position changes from x 1 = 75 m to x 2 = 35 m. What was the runner’s average velocity?

Given x 1 = 75 m, x 2 = 35 m, Δt = 5 s

The average velocity is given by

v avg = (x 2 – x 1 )/Δt = (35 m – 75 m)/5s = – 8 m/s

Therefore, the runner is running in the negative x-direction.

Problem 2 . A commuter train travels from New York to Philadelphia in 1 hour and 25 minutes and from Philadelphia to New York in 1 hour and 35 minutes. The distance between the two stations is approximately 96 miles. What is (a) the average velocity of the train and (b) the average speed of the train in m/s? (1 mile = 1.6 km)

a) The average velocity of the train is zero since the displacement is zero. (The train returns to New York).

d = 96 miles = 96 x 1.6 = 153.6 km = 153.6 x 10 3 m

t 1 = 1 hr 25 mins = 60 mins + 25 mins = 85 mins = 85 x 60 = 5100 s

Therefore, v 1 = 153.6 x 10 3 m / 5100 s = 30.2 m/s

t 2 = 1 hr 35 mins = 60 mins + 35 mins = 95 mins = 95 x 60 = 5700 s

Therefore, v 2 = 153.6 x 10 3 m / 5700 s = 26.9 m/s

The average speed is (30.2 m/s + 26.9 m/s)/2 = 28.55 m/s

• Speed vs. Velocity – Physicsclassroom.com
• What is Velocity? – Khanacademy.org
• What is Velocity in Physics? – Thoughtco.com
• Speed and Velocity – Physics.info
• Instantaneous Velocity and Speed – Pressbooks.online.ucf.edu
• Time, Velocity, and Speed – Phys.libretexts.org
• Velocity – Hyperphysics.phy-astr.gsu.edu

Article was last reviewed on Friday, July 28, 2023

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The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6 . The four kinematic equations are:

In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in v i ) indicates that the velocity value is the initial velocity value and a subscript of f (as in v f ) indicates that the velocity value is the final velocity value.

Problem-Solving Strategy

• Construct an informative diagram of the physical situation.
• Identify and list the given information in variable form.
• Identify and list the unknown information in variable form.
• Identify and list the equation that will be used to determine unknown information from known information.
• Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information.
• Check your answer to insure that it is reasonable and mathematically correct.

The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below.  

Example Problem A

Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s 2 , then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.)

The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the v f value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity ( v i ) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration ( a ) of the car is given as - 8.00 m/s 2 . (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below.  

The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are v f , v i , a , and d . Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables.

Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.

(0 m/s) 2 = (30.0 m/s) 2 + 2 • (-8.00 m/s 2 ) • d

0 m 2 /s 2 = 900 m 2 /s 2 + (-16.0 m/s 2 ) • d

(16.0 m/s 2 ) • d = 900 m 2 /s 2 - 0 m 2 /s 2

(16.0 m/s 2 )*d = 900 m 2 /s 2

d = (900 m 2 /s 2 )/ (16.0 m/s 2 )

The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!  

Example Problem B

Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s 2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period.

Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the v i value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration ( a ) of the car is 6.00 m/s 2 . And the time ( t ) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below.

The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, v i , a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables.  

d = (0 m/s) • (4.1 s) + ½ • (6.00 m/s 2 ) • (4.10 s) 2

d = (0 m) + ½ • (6.00 m/s 2 ) • (16.81 s 2 )

d = 0 m + 50.43 m

The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!

The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6 , we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions.  

Speed and Velocity

Speed is how fast something moves.

Velocity is speed with a direction .

Saying Ariel the Dog runs at 9 km/h (kilometers per hour) is a speed.

But saying he runs 9 km/h Westwards is a velocity.

Imagine something moving back and forth very fast: it has a high speed, but a low (or zero) velocity.

Speed is measured as distance moved over time.

Speed = Distance Time

Example: A car travels 50 km in one hour.

Its average speed is 50 km per hour (50 km/h)

Speed = Distance Time = 50 km 1 hour

We can also use these symbols:

Speed = Δs Δt

Where Δ (" Delta ") means "change in", and

• s means distance ("s" for "space")
• t means time

Example: You run 360 m in 60 seconds.

So your speed is 6 meters per second (6 m/s).

Speed is commonly measured in:

• meters per second (m/s or m s -1 ), or
• kilometers per hour (km/h or km h -1 )

A km is 1000 m, and there are 3600 seconds in an hour, so we can convert like this (see Unit Conversion Method to learn more):

So 1 m/s is equal to 3.6 km/h

Example: What is 20 m/s in km/h ?

20 m/s × 3.6 km/h 1 m/s = 72 km/h

Example: What is 120 km/h in m/s ?

120 km/h × 1 m/s 3.6 km/h = 33.333... m/s

Average vs Instantaneous Speed

The examples so far calculate average speed : how far something travels over a period of time.

But speed can change as time goes by. A car can go faster and slower, maybe even stop at lights.

So there is also instantaneous speed : the speed at an instant in time. We can try to measure it by using a very short span of time (the shorter the better).

Example: Sam uses a stopwatch and measures 1.6 seconds as the car travels between two posts 20 m apart. What is the instantaneous speed ?

Well, we don't know exactly, as the car may have been speeding up or slowing down during that time, but we can estimate:

20 m 1.6 s = 12.5 m/s = 45 km/h

It is really still an average, but is close to an instantaneous speed.

Constant Speed

When the speed does not change it is constant .

For constant speed, the average and instantaneous speeds are the same.

Because the direction is important velocity uses displacement instead of distance:

Velocity = Displacement Time in a direction.

Example: You walk from home to the shop in 100 seconds, what is your speed and what is your velocity?

Speed = 220 m 100 s = 2.2 m/s

Velocity = 130 m 100 s East = 1.3 m/s East

You forgot your money so you turn around and go back home in 120 more seconds: what is your round-trip speed and velocity?

The total time is 100 s + 120 s = 220 s:

Speed = 440 m 220 s = 2.0 m/s

Velocity = 0 m 220 s = 0 m/s

Yes, the velocity is zero as you ended up where you started.

Learn more at Vectors .

Motion is relative. When we say something is "at rest" or "moving at 4 m/s" we forget to say "in relation to me" or "in relation to the ground", etc.

Think about this: are you really standing still? You are on planet Earth which is spinning at 40,075 km per day (about 1675 km/h or 465 m/s), and moving around the Sun at about 100,000 km/h, which is itself moving through the Galaxy.

Next time you are out walking, imagine you are still and it is the world that moves under your feet. Feels great.

It is all relative!

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How to Calculate Velocity

Last Updated: May 2, 2024 Fact Checked

This article was co-authored by Sean Alexander, MS . Sean Alexander is an Academic Tutor specializing in teaching mathematics and physics. Sean is the Owner of Alexander Tutoring, an academic tutoring business that provides personalized studying sessions focused on mathematics and physics. With over 15 years of experience, Sean has worked as a physics and math instructor and tutor for Stanford University, San Francisco State University, and Stanbridge Academy. He holds a BS in Physics from the University of California, Santa Barbara and an MS in Theoretical Physics from San Francisco State University. There are 10 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,536,854 times.

Quick Formulas

• a = acceleration     t = time

Finding Average Velocity

• The car was displaced by (41m - 5m) = 36 meters east.
• The diver ended up 4 meters below the starting point, so her displacement is 4 meters downward, or -4 meters. (0 + 1 - 5 = -4). Even though the diver traveled six meters (one up, then five down), what matters is that the end point is four meters below the start point.

• Example 1 (cont.): The problem tells us that the car took 8 seconds to go from the start point to the end point, so this is the change in time.
• Example 2 (cont.): If the diver jumped at t = 7 seconds and hits the water at t = 8 seconds, the change in time = 8s - 7s = 1 second.

• Draw a diagram and connect the start point and end point with a straight line. This is the hypotenuse of a triangle, so solve for its length of this line using properties of right triangles . In this case, the displacement is 5 meters northeast.
• At some point, your math teacher may require you to find the exact direction traveled (the angle above the horizontal). You can do this by using geometry or by adding vectors.

Finding Velocity from Acceleration

• a = 10 m/s 2
• (a * t) = (10 m/s 2 * 5 s) = 50 m/s increase in velocity.

• In our example, since the ship started going north and did not change direction, its final velocity is 52 m/s north.

Circular Velocity

• The circular velocity of an object is calculated by dividing the circumference of the circular path by the time period over which the object travels.
• v = (2πr) / T
• Note that 2πr equals the circumference of the circular path.
• r stands for "radius"
• T stands for "time period"

• Circumference = 2πr = ~ (2)(3.14)(8 m) = 50.24 m

• The circular velocity of the object is 1.12 m/s.

Calculator, Practice Problems, and Answers

Community Q&A

• Average velocity measures the average velocity an object travels over the full course of its path. Instantaneous velocity measures the velocity of an object at a specific moment along its path. Thanks Helpful 0 Not Helpful 0

You Might Also Like

• ↑ Sean Alexander, MS. Academic Tutor. Expert Interview. 14 May 2020.
• ↑ https://www.khanacademy.org/science/physics/one-dimensional-motion/displacement-velocity-time/a/what-is-velocity
• ↑ https://www.physicsclassroom.com/class/1DKin/Lesson-1/Speed-and-Velocity
• ↑ https://openstax.org/books/university-physics-volume-1/pages/3-1-position-displacement-and-average-velocity
• ↑ http://www.physicsclassroom.com/Class/1DKin/U1L1c.cfm
• ↑ https://www.omnicalculator.com/physics/velocity
• ↑ http://easycalculation.com/physics/classical-physics/circular-velocity.php
• ↑ https://www.omnicalculator.com/physics/angular-velocity
• ↑ https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Trigonometry_(Sundstrom_and_Schlicker)/01%3A_The_Trigonometric_Functions/1.04%3A_Velocity_and_Angular_Velocity
• ↑ https://physics.info/velocity/

About This Article

Velocity is defined as the speed at which an object travels in a given direction. The right formula to use for calculating velocity depends on a few different factors, such as whether the object is accelerating at a constant rate, or whether it is moving in a circle as opposed to a line. The most basic formula for calculating velocity is velocity (v) = distance (d)/time (t). If you don’t already know the time and distance, you’ll need to calculate them first. Subtract the initial position from the final position to find distance, and subtract the start time from the end time to find the time. For instance, if a runner began sprinting due east at the 22-meter mark along a track and ended up at the 52-meter mark, you’d subtract 22 from 52 to find the distance, or displacement, of 30 meters. Similarly, if they began their sprint at 5:35:01 pm and ended it at 5:35:06 pm, you can find the time by subtracting 1 from 6, giving you 5 seconds. This will tell you that they ran 30 meters in 5 seconds, which means that they maintained an average velocity of 6 m/s east. If you’re finding the velocity of an object that’s accelerating instead of moving at a constant rate, things get a little more complicated. If you know the acceleration rate of the object, you can find the final velocity using the formula vf (final velocity) = vi (initial velocity) + a(t) (acceleration x time). For example, if an object accelerated north at a rate of 5m/s2 over 5 seconds and had a starting velocity of 6 m/s, its final velocity would be 6m/s + (5m/s2 x 5s), or 31m/s north. Once you know both the final and initial velocity, you can calculate the average velocity of an accelerating object. To do this, add initial velocity to final velocity and divide the result by 2. In this case, 6m/s + 30m/s divided by 2 = 18 m/s north. The method for finding the velocity of an object around a circle is a little different. To do this, use the formula v (velocity) = 2πr (the circumference of the circle)/t (time). For example, an object that moves around a circle with a radius of 50 meters in 13 seconds would have a velocity of 2π(50)m/13s, or approximately 24.17 m/s. To learn more, such as how to calculate average or circular velocity, keep reading the article! Did this summary help you? Yes No

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Speed, Velocity, and Acceleration Problems

By Abnurlion

On December 26, 2022

In MECHANICS

Table of Contents

What is Speed?

Definition of Speed : Speed is the rate of change of distance with time . Speed is different from velocity because it’s not in a specified direction. In this article, you will learn how to solve speed, velocity, and acceleration problems.

Additionally, you need to know that speed is a scalar quantity and we can write its symbol as S. The formula for calculating the speed of an object is:

Speed, S = Distance (d) / Time (t)

Thus, s = d/t

Note: In most cases, we also use S as a symbol for distance.

The S.I unit for speed is meter per second (m/s) or ms -1

Non-Uniform or Average Speed : This is a non-steady distance covered by an object at a particular period of time. We can also define non-uniform speed as the type of distance that an object covered at an equal interval of time.

The formula for calculating non-uniform speed is

Average speed = Total distance covered by the object / Total time taken

Actual speed: This is also known as the instantaneous speed of an object which is the distance covered by an object over a short interval of time.

You may also like to read:

How to Find Displacement in Physics

also How to Find a Position in Physics

and How to Calculate Bearing in Physics

What is Velocity?

Definition of Velocity: Velocity is the rate of displacement with time. Velocity is the speed of an object in a specified direction. The unit of velocity is the same as that of speed which is meter per second (ms -1 ). We use V as a symbol for velocity.

Note: We often use U to indicate initial speed, and V to indicate final speed.

The formula for calculating Velocity (V) = displacement (S) / time (t)

The difference between velocity and speed is the presence of displacement and distance respectively. Because displacement is a measure of separation in a specified direction, while distance is not in a specified direction. Velocity is a vector quantity.

Uniform Velocity

Definition of uniform velocity: The rate of change of displacement is constant no matter how small the time interval may be. Also, uniform velocity is the distance covered by an object in a specified direction in an equal time interval.

The formula for uniform velocity = Total displacement / Total time taken

What is Acceleration?

Definition of Acceleration: Acceleration is the rate of change of velocity with time. Acceleration is measured in meters per second square (ms -2 ). The symbol for acceleration is a. Acceleration is also a vector quantity.

The formula for acceleration , a = change in velocity (v)/time taken (t)

Thus, a = v/t

We can also write acceleration as

a = change in velocity/time = ΔV/Δt = (v – u)/t

[where v = final velocity, u = initial velocity, and t = time taken]

Uniform Acceleration

In the case of uniform acceleration , the rate of change of velocity with time is constant.

Average velocity of the object = (Initial velocity + final velocity)/2

Average velocity = (v + u)/2

What is Retardation?

Retardation is the decreasing rate of change in velocity moved, covered, or traveled by an object.

The formula for calculating retardation is

Retardation = Change in a decrease in velocity/time taken

Equations of Motion

You can also apply the following equations of motion to calculate the speed, velocity, and acceleration of the body:

• s = [ (v + u)/2 ]t
• v 2 = u 2 + 2as

s = ut + (1/2)at 2

Where v = final velocity, u = initial velocity, t = time, a = acceleration, and s = distance

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Solved Problems of Speed, Velocity, and Acceleration

Here are solved problems to help you understand how to calculate speed, velocity, and acceleration:

A train moves at a speed of 54km/h for a one-quarter minute. Find the distance travelled by train.

From the question above

Speed = 54 km/h = (54 x 1000)/(60 x 60) = 54,000/3,600 = 15 m/s

Time = one quarter minute = 1/4 minute = (1/4) x 60 = 15 seconds

Since we have

speed = distance/time

After cross-multiplication, we will now have

Distance = speed x time

We can now insert our data into the above expression

Distance = 15 m/s x 15 s = 225 m

Therefore, the distance travelled by train is 225 meters.

A car travelled a distance of 5km in 50 seconds. Find the speed in meters per second.

Distance = 5km = 5 x 1000m = 5,000m

Time = 50 seconds

and the formula for speed

speed = distance/time = 5000/50 = 100m/s

A motorcycle starting from rest moves with a uniform acceleration until it attains a speed of 108 kilometres per hour after 15 seconds. Find its acceleration.

Initial velocity, u = 0 (because the motorcycle starts from rest)

Final velocity, v = speed = 108 km/h = (108 x 1000m) / (60 x 60s) = 108,000/3,600 = 30m/s

Time taken, t = 15 seconds

Therefore, we can now apply the formula that says

Acceleration = change in velocity/ time = (v-u)/t = (30-0)/15 =30/15 = 2ms -2

Therefore, the acceleration of the motorcycle is 2ms -2

A bus covers 50 kilometres in 1 hour. What is it is the average speed?

Total distance covered = 50 km = 50 x 1000m = 50,000m

Time taken = 1 hour = 1 x 60 x 60s =3,600s

Therefore, we can now calculate the average speed of the bus by substituting our data into the above formula

Average speed = 50,000/3,600 = 13.9 m/s

Therefore, the average speed of the bus is 13.9m/s or approximately 14 meters per second.

A car travels 80 km in 1 hour and then another 20 km in the next hour. Find the average velocity of the car.

Initial displacement of the car = 80km

Final displacement of the car = 20km

The total displacement of the car = initial displacement of the car + final displacement of the car

The total displacement of the car = 80km + 20km = 100km

The time for 80km is 1hr

And the time for 20km is 1hr

Total time taken = The time for 80km (1hr) + The time for 20km (1hr)

Total time taken = 1hr + 1hr = 2hrs

Now, to calculate the average velocity of the car, we apply the formula that says

Average velocity = total displacement/total time taken = 100km/2hrs = 50km/h

We can further convert the above answer into meters per second

Average velocity = 50km/h = (50 x 1000m)/(1 x 60 x 60s) = 50,000/3,600 = 13.9m/s or 14ms -1

Therefore, the average velocity of the car is 14ms -1

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A body falls from the top of a tower 100 meters high and hits the ground in 5 seconds. Find its acceleration.

Displacement = 100m

Time = 5 seconds

and we can apply the formula for acceleration that says

acceleration, a = Displacement/time 2 = 100/5 2 = 100/25 = 4ms -2

Therefore the acceleration due to the gravity of the body is 4ms -2

Note: The acceleration due to the gravity of a body on the surface of the earth is constant (9.8ms -2 ), even though there may be a slight difference due to the mass and altitude of the body.

An object is thrown vertically upward at an initial velocity of 10ms -1 and reaches a maximum height of 50 meters. Find its initial upward acceleration.

Initial velocity, u = 10ms -1

Final velocity, v = 0

maximum height = displacement = 50m

Initial upward acceleration, a =?

When we apply the formula that says a = (v 2 – u 2 )/2d we will have

a = (0 – 10 2 )/(2 x 50) = -100/100 = -1 ms -2

Hence, since our acceleration is negative, we can now say that we are dealing with retardation or deceleration.

Therefore, the retardation is -1ms -2

Note: Retardation is the negative of acceleration, thus it is written in negative form.

A car is traveling at a velocity of 8ms -1 and experiences an acceleration of 5ms -2 . How far does it travel in 4 seconds?

Initial velocity, u = 8ms -1

acceleration, a = 5ms -2

Distance, s =?

time, t = 4s

We can apply the formula that says

s = 8 x 4 + (1/2) x 5 x 4 2

s = 32 + (1/2) x 80 = 32 + 40 = 72m

Therefore, the distance covered by the car in 4s is 72 meters.

A body is traveling at a velocity of 10m/s and experiences a deceleration of 5ms -2 . How long does it take the body to come to a complete stop?

Initial velocity, u = 10m/s

acceleration , a = retardation = -5ms -2

time, t = ?

We already know that acceleration, a = change in velocity/time

This implies that

Time, t = change in velocity/acceleration

t = (v – u)/t = (0-10)/-5 = -10/-5 = 2s

Therefore, the time it takes the car to stop is 2 seconds .

A body is traveling at a velocity of 20 m/s and has a mass of 10 kg. How much force is required to change its velocity by 10 m/s in 5 seconds?

Change in velocity, v =10 m/s

mass of the object, m = 10kg

time, t = 5s

We can apply newton’s second law of motion which says f = ma

and since a = change in velocity/time

we will have an acceleration equal to

a = 10/5 = 2ms -2

Therefore, to find the force, we can now say

f = ma = 10 x 2 = 20N

Therefore, the force that can help us to change the velocity by 10 m/s in 5 seconds is 20-Newton.

Drop a comment if there is anything you don’t understand about speed, velocity, and acceleration Problems.

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Solved Speed, Velocity, and Acceleration Problems

Simple problems on speed, velocity, and acceleration with descriptive answers are presented for the AP Physics 1 exam and college students. In each solution, you can find a brief tutorial. 

Speed and velocity Problems: 

Problem (1): What is the speed of a rocket that travels $8000\,{\rm m}$ in $13\,{\rm s}$?

Solution : Speed is defined in physics  as the total distance divided by the elapsed time,  so the rocket's speed is \[\text{speed}=\frac{8000}{13}=615.38\,{\rm m/s}\]

Problem (2): How long will it take if you travel $400\,{\rm km}$ with an average speed of $100\,{\rm m/s}$?

Solution : Average speed is the ratio of the total distance to the total time. Thus, the elapsed time is \begin{align*} t&=\frac{\text{total distance}}{\text{average speed}}\\ \\ &=\frac{400\times 10^{3}\,{\rm m}}{100\,{\rm m/s}}\\ \\ &=4000\,{\rm s}\end{align*} To convert it to hours, it must be divided by $3600\,{\rm s}$ which gives $t=1.11\,{\rm h}$.

Problem (3): A person walks $100\,{\rm m}$ in $5$ minutes, then $200\,{\rm m}$ in $7$ minutes, and finally $50\,{\rm m}$ in $4$ minutes. Find its average speed. 

Solution : First find its total distance traveled ($D$) by summing all distances in each section, which gets $D=100+200+50=350\,{\rm m}$. Now, by definition of average speed, divide it by the total time elapsed $T=5+7+4=16$ minutes.

But keep in mind that since the distance is in SI units, so the time traveled must also be in SI units, which is $\rm s$. Therefore, we have\begin{align*}\text{average speed}&=\frac{\text{total distance} }{\text{total time} }\\ \\ &=\frac{350\,{\rm m}}{16\times 60\,{\rm s}}\\ \\&=0.36\,{\rm m/s}\end{align*}

Problem (4): A person walks $750\,{\rm m}$ due north, then $250\,{\rm m}$ due east. If the entire walk takes $12$ minutes, find the person's average velocity. 

Solution : Average velocity , $\bar{v}=\frac{\Delta x}{\Delta t}$, is displacement divided by the elapsed time. Displacement is also a vector that obeys the addition vector rules. Thus, in this velocity problem, add each displacement to get the total displacement . 

In the first part, displacement is $\Delta x_1=750\,\hat{j}$ (due north) and in the second part $\Delta x_2=250\,\hat{i}$ (due east). The total displacement vector is $\Delta x=\Delta x_1+\Delta x_2=750\,\hat{i}+250\,\hat{j}$ with magnitude of  \begin{align*}|\Delta x|&=\sqrt{(750)^{2}+(250)^{2}}\\ \\&=790.5\,{\rm m}\end{align*} In addition, the total elapsed time is $t=12\times 60$ seconds. Therefore, the magnitude of the average velocity is \[\bar{v}=\frac{790.5}{12\times 60}=1.09\,{\rm m/s}\]

Problem (5): An object moves along a straight line. First, it travels at a velocity of $12\,{\rm m/s}$ for $5\,{\rm s}$ and then continues in the same direction with $20\,{\rm m/s}$ for $3\,{\rm s}$. What is its average speed?

Solution: Average velocity is displacement divided by elapsed time, i.e., $\bar{v}\equiv \frac{\Delta x_{tot}}{\Delta t_{tot}}$.

Here, the object goes through two stages with two different displacements, so add them to find the total displacement. Thus,\[\bar{v}=\frac{x_1 + x_2}{t_1 +t_2}\] Again, to find the displacement, we use the same equation as the average velocity formula, i.e., $x=vt$. Thus, displacements are obtained as $x_1=v_1\,t_1=12\times 5=60\,{\rm m}$ and $x_2=v_2\,t_2=20\times 3=60\,{\rm m}$. Therefore, we have \begin{align*} \bar{v}&=\frac{x_1+x_2}{t_1+t_2}\\ \\&=\frac{60+60}{5+3}\\ \\&=\boxed{15\,{\rm m/s}}\end{align*}

Problem (6): A plane flies the distance between two cities in $1$ hour and $30$ minutes with a velocity of $900\,{\rm km/h}$. Another plane covers that distance at $600\,{\rm km/h}$. What is the flight time of the second plane?

Solution: first find the distance between two cities using the average velocity formula $\bar{v}=\frac{\Delta x}{\Delta t}$ as below \begin{align*} x&=vt\\&=900\times 1.5\\&=1350\,{\rm km}\end{align*} where we wrote one hour and a half minutes as $1.5\,\rm h$. Now use again the same kinematic equation above to find the time required for another plane \begin{align*} t&=\frac xv\\ \\ &=\frac{1350\,\rm km}{600\,\rm km/h}\\ \\&=2.25\,{\rm h}\end{align*} Thus, the time for the second plane is $2$ hours and $0.25$ of an hour, which converts to minutes as $2$ hours and ($0.25\times 60=15$) minutes.

Problem (7): To reach a park located south of his jogging path, Henry runs along a 15-kilometer route. If he completes the journey in 1.5 hours, determine his speed and velocity.

Solution:  Henry travels his route to the park without changing direction along a straight line. Therefore, the total distance traveled in one direction equals the displacement, i.e, \[\text{distance traveled}=\Delta x=15\,\rm km\]Velocity is displacement divided by the time of travel \begin{align*} \text{velocity}&=\frac{\text{displacement}}{\text{time of travel}} \\\\ &=\frac{15\,\rm km}{1.5\,\rm h} \\\\ &=\boxed{10\,\rm km/h}\end{align*} and by definition, its average speed is \begin{align*} \text{speed}&=\frac{\text{distance covered}}{\text{time interval}}\\\\&=\frac{15\,\rm km}{1.5\,\rm h}\\\\&=\boxed{10\,\rm km/h}\end{align*} Thus, Henry's velocity is $10\,\rm km/h$ to the south, and its speed is $10\,\rm km/h$. As you can see, speed is simply a positive number, with units but velocity specifies the direction in which the object is moving. 

Problem (8): In 15 seconds, a football player covers the distance from his team's goal line to the opposing team's goal line and back to the midway point of the field having 100-yard-length. Find, (a) his average speed, and (b) the magnitude of the average velocity.

Solution:  The total length of the football field is $100$ yards or in meters, $L=91.44\,\rm m$. Going from one goal's line to the other and back to the midpoint of the field takes $15\,\rm s$ and covers a distance of $D=100+50=150\,\rm yd$. 

Distance divided by the time of travel gets the average speed, \[\text{speed}=\frac{150\times 0.91}{15}=9.1\,\rm m/s\] To find the average velocity, we must find the displacement of the player between the initial and final points. 

The initial point is her own goal line and her final position is the midpoint of the field, so she has displaced a distance of $\Delta x=50\,\rm yd$ or $\Delta x=50\times 0.91=45.5\,\rm m$. Therefore, her velocity is calculated as follows \begin{align*} \text{velocity}&=\frac{\text{displacement}}{\text{time elapsed}} \\\\ &=\frac{45.5\,\rm m}{15\,\rm s} \\\\&=\boxed{3.03\quad \rm m/s}\end{align*} Contrary to the previous problem, here the motion is not in one direction, hence, the displacement is not equal to the distance traveled. Accordingly, the average speed is not equal to the magnitude of the average velocity.

Problem (9): You begin at a pillar and run towards the east (the positive $x$ direction) for $250\,\rm m$ at an average speed of $5\,\rm m/s$. After that, you run towards the west for $300\,\rm m$ at an average speed of $4\,\rm m/s$ until you reach a post. Calculate (a) your average speed from pillar to post, and (b) your average velocity from pillar to post. 

Solution : First, you traveled a distance of $L_1=250\,\rm m$ toward east (or $+x$ direction) at $5\,\rm m/s$. Time of travel in this route is obtained as follows \begin{align*} t_1&=\frac{L_1}{v_1}\\\\ &=\frac{250}{5}\\\\&=50\,\rm s\end{align*} Likewise, traveling a distance of $L_2=300\,\rm m$ at $v_2=4\,\rm m/s$ takes \[t_2=\frac{300}{4}=75\,\rm s\]  (a) Average speed is defined as the distance traveled (or path length) divided by the total time of travel \begin{align*} v&=\frac{\text{path length}}{\text{time of travel}} \\\\ &=\frac{L_1+L_2}{t_1+t_2}\\\\&=\frac{250+300}{50+75} \\\\&=4.4\,\rm m/s\end{align*} Therefore, you travel between these two pillars in $125\,\rm s$ and with an average speed of $4.4\,\rm m/s$. 

(b) Average velocity requires finding the displacement between those two points. In the first case, you move $250\,\rm m$ toward $+x$ direction, i.e., $L_1=+250\,\rm m$. Similarly, on the way back, you move $300\,\rm m$ toward the west ($-x$ direction) or $L_2=-300\,\rm m$. Adding these two gives us the total displacement between the initial point and the final point, \begin{align*} L&=L_1+L_2 \\\\&=(+250)+(-300) \\\\ &=-50\,\rm m\end{align*} The minus sign indicates that you are generally displaced toward the west. 

Finally, the average velocity is obtained as follows: \begin{align*} \text{average velocity}&=\frac{\text{displacement}}{\text{time of travel}} \\\\ &=\frac{-50}{125} \\\\&=-0.4\,\rm m/s\end{align*} A negative average velocity indicating motion to the left along the $x$-axis. 

This speed problem better makes it clear to us the difference between average speed and average speed. Unlike average speed, which is always a positive number, the average velocity in a straight line can be either positive or negative. 

Problem (10): What is the average speed for the round trip of a car moving uphill at 40 km/h and then back downhill at 60 km/h? 

Solution : Assuming the length of the hill to be $L$, the total distance traveled during this round trip is $2L$ since $L_{up}=L_{down}=L$. However, the time taken for going uphill and downhill was not provided. We can write them in terms of the hill's length $L$ as $t=\frac L v$. 

Applying the definition of average speed gives us \begin{align*} v&=\frac{\text{distance traveled}}{\text{total time}} \\\\ &=\frac{L_{up}+L_{down}}{t_{up}+t_{down}} \\\\ &=\cfrac{2L}{\cfrac{L}{v_{up}}+\cfrac{L}{v_{down}}} \end{align*} By reorganizing this expression, we obtain a formula that is useful for solving similar problems in the AP Physics 1 exams. \[\text{average speed}=\frac{2v_{up} \times v_{down}}{v_{up}+v_{down}}\] Substituting the numerical values into this, yields \begin{align*} v&=\frac{2(40\times 60)}{40+60} \\\\ &=\boxed{48\,\rm m/s}\end{align*} What if we were asked for the average velocity instead? During this round trip, the car returns to its original position, and thus its displacement, which defines the average velocity, is zero. Therefore, \[\text{average velocity}=0\,\rm m/s\]

Acceleration Problems

Problem (9): A car moves from rest to a speed of $45\,\rm m/s$ in a time interval of $15\,\rm s$. At what rate does the car accelerate? 

Solution : The car is initially at rest, $v_1=0$, and finally reaches $v_2=45\,\rm m/s$ in a time interval $\Delta t=15\,\rm s$. Average acceleration is the change in velocity, $\Delta v=v_2-v_1$, divided by the elapsed time $\Delta t$, so \[\bar{a}=\frac{45-0}{15}=\boxed{3\,\rm m/s^2} \] 

Problem (10): A car moving at a velocity of $15\,{\rm m/s}$, uniformly slows down. It comes to a complete stop in $10\,{\rm s}$. What is its acceleration?

Solution:  Let the car's uniform velocity be $v_1$ and its final velocity $v_2=0$.   Average acceleration is the difference in velocities divided by the time taken, so we have: \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{0-15}{10}\\\\ &=\boxed{-1.5\,{\rm m/s^2}}\end{align*}The minus sign indicates the direction of the acceleration vector, which is toward the $-x$ direction.

Problem (11): A car moves from rest to a speed of $72\,{\rm km/h}$ in $4\,{\rm s}$. Find the acceleration of the car.

Solution: Known: $v_1=0$, $v_2=72\,{\rm km/h}$, $\Delta t=4\,{\rm s}$.  Average acceleration is defined as the difference in velocities divided by the time interval between those points \begin{align*}\bar{a}&=\frac{v_2-v_1}{t_2-t_1}\\\\&=\frac{20-0}{4}\\\\&=5\,{\rm m/s^2}\end{align*} In above, we converted $\rm km/h$ to the SI unit of velocity ($\rm m/s$) as \[1\,\frac{km}{h}=\frac {1000\,m}{3600\,s}=\frac{10}{36}\, \rm m/s\] so we get \[72\,\rm km/h=72\times \frac{10}{36}=20\,\rm m/s\] 

Problem (12): A race car accelerates from an initial velocity of $v_i=10\,{\rm m/s}$ to a final velocity of $v_f = 30\,{\rm m/s}$ in a time interval of $2\,{\rm s}$. Determine its average acceleration.

Solution:  A change in the velocity of an object $\Delta v$ over a time interval $\Delta t$ is defined as an average acceleration. Known: $v_i=10\,{\rm m/s}$, $v_f = 30\,{\rm m/s}$, $\Delta t=2\,{\rm s}$. Applying definition of average acceleration, we get \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\&=\frac{30-10}{2}\\&=10\,{\rm m/s^2}\end{align*}

Problem (13): A motorcycle starts its trip along a straight line with a velocity of $10\,{\rm m/s}$ and ends with $20\,{\rm m/s}$ in the opposite direction in a time interval of $2\,{\rm s}$. What is the average acceleration of the car?

Solution:  Known: $v_i=10\,{\rm m/s}$, $v_f=-20\,{\rm m/s}$, $\Delta t=2\,{\rm s}$, $\bar{a}=?$. Using average acceleration definition we have \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\\\&=\frac{(-20)-10}{2}\\\\ &=\boxed{-15\,{\rm m/s^2}}\end{align*}Recall that in the definition above, velocities are vector quantities. The final velocity is in the opposite direction from the initial velocity so a negative must be included.

Problem (14): A ball is thrown vertically up into the air by a boy. After $4$ seconds, it reaches the highest point of its path. How fast does the ball leave the boy's hand?

Solution : At the highest point, the ball has zero speed, $v_2=0$. It takes the ball $4\,\rm s$ to reach that point. In this problem, our unknown is the initial speed of the ball, $v_1=?$. Here, the ball accelerates at a constant rate of $g=-9.8\,\rm m/s^2$ in the presence of gravity.

When the ball is tossed upward, the only external force that acts on it is the gravity force. 

Using the average acceleration formula $\bar{a}=\frac{\Delta v}{\Delta t}$ and substituting the numerical values into this, we will have \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{0-v_1}{4} \\\\ \Rightarrow \boxed{v_1=39.2\,\rm m/s} \end{gather*} Note that $\Delta v=v_2-v_1$. 

Problem (15): A child drops crumpled paper from a window. The paper hit the ground in $3\,\rm s$. What is the velocity of the crumpled paper just before it strikes the ground? 

Solution : The crumpled paper is initially in the child's hand, so $v_1=0$. Let its speed just before striking be $v_2$. In this case, we have an object accelerating down in the presence of gravitational force at a constant rate of $g=-9.8\,\rm m/s^2$. Using the definition of average acceleration, we can find $v_2$ as below \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{v_2-0}{3} \\\\ \Rightarrow v_2=3\times (-9.8)=\boxed{-29.4\,\rm m/s} \end{gather*} The negative shows us that the velocity must be downward, as expected!

Problem (16): A car travels along the $x$-axis for $4\,{\rm s}$ at an average velocity of $10\,{\rm m/s}$ and $2\,{\rm s}$ with an average velocity of $30\,{\rm m/s}$ and finally $4\,{\rm s}$ with an average velocity $25\,{\rm m/s}$. What is its average velocity across the whole path?

Solution: There are three different parts with different average velocities. Assume each trip is done in one dimension without changing direction. Thus, displacements associated with each segment are the same as the distance traveled in that direction and is calculated as below: \begin{align*}\Delta x_1&=v_1\,\Delta t_1\\&=10\times 4=40\,{\rm m}\\ \\ \Delta x_2&=v_2\,\Delta t_2\\&=30\times 2=60\,{\rm m}\\ \\ \Delta x_3&=v_3\,\Delta t_3\\&=25\times 4=100\,{\rm m}\end{align*}Now use the definition of average velocity, $\bar{v}=\frac{\Delta x_{tot}}{\Delta t_{tot}}$, to find it over the whole path\begin{align*}\bar{v}&=\frac{\Delta x_{tot}}{\Delta t_{tot}}\\ \\&=\frac{\Delta x_1+\Delta x_2+\Delta x_3}{\Delta t_1+\Delta t_2+\Delta t_3}\\ \\&=\frac{40+60+100}{4+2+4}\\ \\ &=\boxed{20\,{\rm m/s}}\end{align*}

Problem (17): An object moving along a straight-line path. It travels with an average velocity $2\,{\rm m/s}$ for $20\,{\rm s}$ and $12\,{\rm m/s}$ for $t$ seconds. If the total average velocity across the whole path is $10\,{\rm m/s}$, then find the unknown time $t$.

Solution: In this velocity problem, the whole path $\Delta x$ is divided into two parts $\Delta x_1$ and $\Delta x_2$ with different average velocities and times elapsed, so the total average velocity across the whole path is obtained as \begin{align*}\bar{v}&=\frac{\Delta x}{\Delta t}\\\\&=\frac{\Delta x_1+\Delta x_2}{\Delta t_1+\Delta t_2}\\\\&=\frac{\bar{v}_1\,t_1+\bar{v}_2\,t_2}{t_1+t_2}\\\\10&=\frac{2\times 20+12\times t}{20+t}\\\Rightarrow t&=80\,{\rm s}\end{align*}

Note : whenever a moving object, covers distances $x_1,x_2,x_3,\cdots$ in $t_1,t_2,t_3,\cdots$ with constant or average velocities $v_1,v_2,v_3,\cdots$ along a straight-line without changing its direction, then its total average velocity across the whole path is obtained by one of the following formulas

• Distances and times are known:\[\bar{v}=\frac{x_1+x_2+x_3+\cdots}{t_1+t_2+t_3+\cdots}\]
• Velocities and times are known: \[\bar{v}=\frac{v_1\,t_1+v_2\,t_2+v_3\,t_3+\cdots}{t_1+t_2+t_3+\cdots}\]
• Distances and velocities are known:\[\bar{v}=\frac{x_1+x_2+x_3+\cdots}{\frac{x_1}{v_1}+\frac{x_2}{v_2}+\frac{x_3}{v_3}+\cdots}\]

Problem (18): A car travels one-fourth of its path with a constant velocity of $10\,{\rm m/s}$, and the remaining with a constant velocity of $v_2$. If the total average velocity across the whole path is $16\,{\rm m/s}$, then find the $v_2$?

Solution: This is the third case of the preceding note. Let the length of the path be $L$ so \begin{align*}\bar{v}&=\frac{x_1+x_2}{\frac{x_1}{v_1}+\frac{x_2}{v_2}}\\\\16&=\frac{\frac 14\,L+\frac 34\,L}{\frac{\frac 14\,L}{10}+\frac{\frac 34\,L}{v_2}}\\\\\Rightarrow v_2&=20\,{\rm m/s}\end{align*}

Problem (19): An object moves along a straight-line path. It travels for $t_1$ seconds with an average velocity $50\,{\rm m/s}$ and $t_2$ seconds with a constant velocity of $25\,{\rm m/s}$. If the total average velocity across the whole path is $30\,{\rm m/s}$, then find the ratio $\frac{t_2}{t_1}$?

Solution: the velocities and times are known, so we have \begin{align*}\bar{v}&=\frac{v_1\,t_1+v_2\,t_2}{t_1+t_2}\\\\30&=\frac{50\,t_1+25\,t_2}{t_1+t_2}\\\\ \Rightarrow \frac{t_2}{t_1}&=4\end{align*} 

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In the following section, some sample AP Physics 1 problems on acceleration are provided.

Problem (20): An object moves with constant acceleration along a straight line. If its velocity at instant of $t_1 = 3\,{\rm s}$ is $10\,{\rm m/s}$ and at the moment of $t_2 = 8\,{\rm s}$ is $20\,{\rm m/s}$, then what is its initial speed?

Solution: Let the initial speed at time $t=0$ be $v_0$. Now apply average acceleration definition in the time intervals $[t_0,t_1]$ and $[t_0,t_2]$ and equate them.\begin{align*}\text{average acceleration}\ \bar{a}&=\frac{\Delta v}{\Delta t}\\\\\frac{v_1 - v_0}{t_1-t_0}&=\frac{v_2-v_0}{t_2-t_0}\\\\ \frac{10-v_0}{3-0}&=\frac{20-v_0}{8-0}\\\\ \Rightarrow v_0 &=4\,{\rm m/s}\end{align*} In the above, $v_1$ and $v_2$ are the velocities at moments $t_1$ and $t_2$, respectively. 

Problem (21): For $10\,{\rm s}$, the velocity of a car that travels with a constant acceleration, changes from $10\,{\rm m/s}$ to $30\,{\rm m/s}$. How far does the car travel?

Solution: Known: $\Delta t=10\,{\rm s}$, $v_1=10\,{\rm m/s}$ and $v_2=30\,{\rm m/s}$. 

Method (I) Without computing the acceleration: Recall that in the case of constant acceleration, we have the following kinematic equations for average velocity and displacement:\begin{align*}\text{average velocity}:\,\bar{v}&=\frac{v_1+v_2}{2}\\\text{displacement}:\,\Delta x&=\frac{v_1+v_2}{2}\times \Delta t\\\end{align*}where $v_1$ and $v_2$ are the velocities in a given time interval. Now we have \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\\&=\frac{10+30}{2}\times 10\\&=200\,{\rm m}\end{align*}

Method (II) with computing acceleration: Using the definition of average acceleration, first determine it as below \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{30-10}{10}\\\\&=2\,{\rm m/s^2}\end{align*} Since the velocities at the initial and final points of the problem are given so use the below time-independent kinematic equation to find the required displacement \begin{align*} v_2^{2}-v_1^{2}&=2\,a\Delta x\\\\ (30)^{2}-(10)^{2}&=2(2)\,\Delta x\\\\ \Rightarrow \Delta x&=\boxed{200\,{\rm m}}\end{align*}

Problem (22): A car travels along a straight line with uniform acceleration. If its velocity at the instant of $t_1=2\,{\rm s}$ is $36\,{\rm km/s}$ and at the moment $t_2=6\,{\rm s}$ is $72\,{\rm km/h}$, then find its initial velocity (at $t_0=0$)?

Solution: Use the equality of definition of average acceleration $a=\frac{v_f-v_i}{t_f-t_i}$ in the time intervals $[t_0,t_1]$ and $[t_0,t_2]$ to find the initial velocity as below \begin{align*}\frac{v_2-v_0}{t_2-t_0}&=\frac{v_1-v_0}{t_1-t_0}\\\\ \frac{20-v_0}{6-0}&=\frac{10-v_0}{2-0}\\\\ \Rightarrow v_0&=\boxed{5\,{\rm m/s}}\end{align*}

All these kinematic problems on speed, velocity, and acceleration are easily solved by choosing an appropriate kinematic equation. Keep in mind that these motion problems in one dimension are of the uniform or constant acceleration type. Projectiles are also another type of motion in two dimensions with constant acceleration.

Author:   Dr. Ali Nemati

Date Published: 9/6/2020

Updated: Jun 28,  2023

© 2015 All rights reserved. by Physexams.com

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Precalculus : Solve Linear Velocity Problems

Study concepts, example questions & explanations for precalculus, all precalculus resources, example questions, example question #1 : solve linear velocity problems.

None of the other answers

Since the plane is traveling NE, we know that it is traveling at an angle of 45 degrees north of east. Therefore, we can represent this in a diagram:

We have all of the information we need to solve for the velocity of the plane. We have an angle and a side adjacent to that angle. Therefore, we can use the cosine function to solve for the velocity.

Rearranging for v, we get:

Example Question #1 : Angular And Linear Velocity

Example Question #3 : Angular And Linear Velocity

An object is moving in a circular path. How will its linear velocity change if the diameter of the circular path was decreased by one half?

Write the formula for the linear velocity.

The diameter is twice the radius.

If the diameter was decreased by one half, the radius will also decrease by one half.

The linear velocity will decrease by one half.

Write the linear velocity formula.

Since the answer is required in meters per second, two conversions will need to be made.

Convert 30 kilometers to meters.

Convert 2 hours to seconds.

Apply the formula.

Example Question #5 : Angular And Linear Velocity

None of the other answers.

And how long does it take for the hand to go around? Linear speed of the clock second hand is 

A clock has a second hand measuring 12cm. What is the linear speed of the tip of the second hand?

None of these/

Linear speed is equal to the arc length traversed divided by the time. We use s to denote arc length.

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Projectile Motion & Quadratic Equations

Projectiles General Word Problems Max/Min Problems

Calculus involves lots of finding maximums, minimums, and zeroes. You will get a taste of that with quadratic word problems. In fact, if/when you reach calculus, you will discover that some of the homework exercises will be identical to those you're doing now; it's just that you'll have new tools for finding the answers.

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A major category of quadratic-equation word problems relates to what is called projectile motion. For our purposes, a projectile is any object that is thrown, shot, or dropped. Almost always, in this context, the object is initially moving directly up or straight down. (If it starts by going up then, naturally, it will later be coming back down.) This initial movement speed is the velocity.

What does the velocity's sign say about the object?

The initial velocity of the object, in these exercises, tells us how the object was released. The initial value of the velocity will be either zero (so the object was just dropped), positive (so it was thrown or shot upward), or negative (so the object was thrown downward).

What does " g " stand for?

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In projectile-motion exercises, the coefficient on the squared term is −½  g . The g stands for the constant of gravity (on Earth), which is −9.8 meters per second square (that is meters per second per second) in metric terms, or −32 feet per second squared in Imperial terms. The "minus" signs reflect the fact that Earth's gravity pulls us, and the object in question, downward.

What does "per second squared" mean?

Acceleration (being the change in speed, rather than the speed itself) is measured in terms of how much the velocity changes per unit time. So, if the velocity of an object is measured in feet per second, then that object's acceleration says how much that velocity changes per unit time; that is, acceleration measures how much the feet per second changes per second. And this duplicate "per second" is how we get "second squared". It's from the physics of the situation.

Which value should I use for gravity?

If a projectile-motion exercise is stated in terms of feet, miles, or some other Imperial unit, then use −32 for gravity; if the units are meters, centimeters, or some other metric unit, then use −9.8 for gravity.

What is the projectile-motion equation?

The projectile-motion equation is s ( t ) = −½  g x 2 +  v 0 x  +  h 0 , where g is the constant of gravity, v 0 is the initial velocity (that is, the velocity at time t  = 0 ), and h 0 is the initial height of the object (that is, the height at of the object at t  = 0 , the time of release).

Yes, you'll need to keep track of all of this stuff when working with projectile motion.

• An object is launched at 19.6 meters per second (m/s) from a 58.8 -meter tall platform. The equation for the object's height s at time t seconds after launch is s ( t ) = −4.9 t 2 + 19.6 t  + 58.8 , where s is in meters. When does the object strike the ground?

What is the height (above ground level) when the object smacks into the ground? Well, zero, obviously. So I'm looking for the time when the height is s  = 0 . I'll set s equal to zero, and solve:

0 = −4.9 t 2 + 19.6 t + 58.8 0 = t 2 − 4 t − 12 0 = ( t − 6)( t + 2)

Then t = 6 or t  = −2 . The second solution is from two seconds before launch, which doesn't make sense in this context. (It makes sense on the graph, because the line crosses the x -axis at −2 , but negative time won't work in this word problem.) So " t  = −2 " is an extraneous solution, and I'll ignore it.

Instead, my answer to this exercise (that is, the correct answer in context) is the other solution value. They asked me for the time, and time here is measured in seconds, so my answer is:

The object strikes the ground six seconds after launch.

Note the construction of the height equation in the problem above. (Yes, we went over this at the beginning, but you're really gonna need this info, so we're revisiting.)

The initial launch height was 58.8 meters, and the constant term was " 58.8 ". The initial velocity (or launch speed) was 19.6 m/s, and the coefficient on the linear term was " 19.6 ". This is always true for these up/down projectile motion problems. (If you have an exercise with sideways motion, the equation will have a different form, but they'll always give you that equation.) The initial velocity is the coefficient for the middle term, and the initial height is the constant term.

And the coefficient on the leading term comes from the force of gravity. This coefficient is negative, since gravity pulls downward, and the value will either be " −4.9 " (if your units are "meters") or " −16 " (if your units are "feet"). Yes, these values are half of the values listed for the gravity constant at the beginning of this page; they've had the ½ multiplied through.

In general, the projectile-motion equation's format is:

s ( t ) = − gt 2 + v 0 t + h 0

...where " g " here is the " 4.9 " or the " 16 " derived from the value of the force of gravity (technically, it's the force of gravity on Earth ), " v 0 " ("vee-naught", or "vee-sub-zero") is the initial velocity, and " h 0 " ("aitch-naught", or "aitch-sub-zero") is the initial height.

Memorize this equation (or at least its meaning), because you may need to know this on the test.

• An object in launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. What will be the object's maximum height? When will it attain this height?

Hmm... They didn't give me the equation this time. But that's okay, because I can create the equation from the information that they did give me. The initial height is 80 feet above ground and the initial speed is 64 ft/s upward. Since my units are feet, then the number for gravity will be −16 , and my equation is:

s ( t ) = −16 t 2 + 64 t + 80

They want me to find the maximum height. For a negative quadratic like this, the maximum will be at the vertex of the upside-down parabola. So they really want me to find the vertex. From graphing , I know how to find the vertex; in this case, the vertex is at ( h ,  k ) = (2, 144):

h = − b / 2 a = −(64)/2(−16) = −64/−32 = 2 k = s (2) = −16(2) 2 + 64(2) + 80 = −16(4) + 128 + 80 = 208 − 64 = 144

But what do the coordinates of this vertex tell me? According to my equation, I'm plugging in time values and extracting height values, so the input 2 was the time and the output 144 is the height.

It takes two seconds to reach the maximum height of 144 feet.

As long as you label clearly, you don't need a complete sentence (like I used above) for your hand-in answer. So you could also give you answer as, "time:  2  secs; height:  144  ft".

• An object is launched from ground level directly upward at 39.2 m/s. For how long is the object at or above a height of 34.3 meters?

My units this time are "meters", so the gravity number will be −4.9 . Since the object started at ground level, the initial height was 0 . Then my equation is:

s ( t ) = −4.9 t 2 + 39.2 t

Since this is a negative quadratic, the graph is an upside-down parabola. I can find the two times when the object is exactly 34.3 meters high, and I know that the object will be above 34.3 meters the whole time in between.

Why "two times", and how do I know that the time period is between those two times? Because the first time will be when the object passes a height of 34.3 meters on its way up to its maximum height, and the second time when be when it passes 34.3 meters as it is falling back down to the ground. So I have to solve the following:

−4.9 t 2 + 39.2 t = 34.3 t 2 − 8 t + 7 = 0 ( t − 7)( t − 1) = 0

The two solutions are at times t  = 1 and t  = 7 . So the object is at 34.3  meters at one second after launch (going up) and againt at seven seconds after launch (coming back down). Subtracting to find the difference, I find that:

The object is at or above 34.3  meters for six seconds.

Again, you don't technically need a complete sentence for your hand-in answer; saying " 6 secs" is probably good enough. But definitely do include the unit "seconds" on your answer.

Don't be surprised if many of your exercises work out as "neatly" as the above examples have. Many textbooks still engineer their exercises carefully, so that you can solve by factoring (that is, by quickly doing the algebra).

However (fair warning!), heavy dependence on calculators is leading more texts to create "interesting" (that is to say, needlessly complicated) exercises, so some (or all) of your exercises may involve much more messy computations than have been displayed here. If so, study these "neat" examples carefully, until you are quite sure you follow the reasoning.

• After the semester is over, you discover that the math department has changed textbooks (again) so the bookstore won't buy back your nearly-new book. You and your friend Herman decide to get creative. You go to the roof of a twelve-story building and look over the edge to the reflecting pool 160  feet below. You drop your book over the edge at the same instant that Herman chucks his book straight down at 48  feet per second. By how many seconds does his book beat yours into the water? (Round your answer to two decimal places.)

Our initial launch heights will be the same: we're both launching from 160  feet above ground. And the gravity number, since we're working in feet, will be −16 . My initial velocity is zero, since I just dropped my book, but my buddy Herman's velocity is a negative 48 , the negative coming from the fact that he chucked his book down rather than up. So our "height" equations are:

mine: s ( t ) = −16 t 2 + 160 his: s ( t ) = −16 t 2 − 48 t + 160

In each case, I need to find the time for the books to reach a height of zero (a height of zero being "ground level"), so:

mine: 0 = −16 t 2 + 160

t 2 − 10 = 0

his: 0 = −16 t 2 − 48 t + 160

t 2 + 3 t − 10 = 0

( t + 5)( t − 2) = 0

t = −5, +2

I will ignore the negative time values as being irrelevant in this context. (There may be a case, at some point, in some exercise, where a negative value does happen to be useful within a given context, so I do need to remember to think about the values and the context, rather than reflexively discarding any and all negative solutions.)

Herman's book hits the water about 1.16 seconds sooner than mine does.

Every once in a while, they'll get clever and put a "projectile" problem into a different environment. The equation will remain the same in structure, but you may have to account for a different value for gravity.

• The International Space Agency has finally landed a robotic explorer on an extra-solar planet. Some probes are extended from the lander's body to conduct various tests. To demonstrate the crushing weight of gravity on this planet, the lander's camera is aimed at a probe's ground-level ejection port, and the port launches a baseball directly upwards at 147 feet per second (ft/s), about the top speed of a professional pitcher. The force due to gravity on this planet is 98  ft/s 2 . Assuming no winds and that the probe can scurry out of the way in time, how long will it take for the ball to smack back into the surface?

To set up my equation for this exercise, I need to keep in mind that the value of the coefficient g from the projectile-motion equation above is one-half of the value of the force due to gravity in a given locale.

In physics, there is the "universal gravitational constant " G , being the gravitational pull inherent to our universe (or at least our region of it). Every object exerts its own gravitational force, which is related to its own mass and the universal constant G . In the "projectile motion" formula, the " g " is half of the value of the gravitational force for that particular body. For instance, the gravitational force on Earth is a downward 32 ft/s 2 , but we used 16 in the equation.

So g for my equation this time will by 98 ÷ 2 = 49 feet per second squared. Then:

s = −49 t 2 + 147 t 0 = −49 t 2 + 147 t 0 = t 2 − 3 t 0 = t  ( t − 3)

Then t = 0 or t = 3 . The first solution represents when the ball was launched, so the second solution is the one I want.

It takes three seconds for the ball to hit the ground.

Note: On Earth, it would take a little over nine seconds for the ball to fall back to the ground.

URL: https://www.purplemath.com/modules/quadprob.htm

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Mechanics (Essentials) - Class 11th

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Impeller Design: Types, Applications, and Simulation

Samir Jaber

published on

May 15th, 2024

approx reading time

An impeller is a rotating component designed to transfer energy from the motor to the fluid, increasing its velocity and pressure as it moves through the machine. A good impeller design ensures optimal fluid dynamics, minimizes energy losses, and contributes to the longevity of the turbomachinery equipment.

In this article, we will discuss the details of impeller design, its challenges, and their solutions. We will also examine how engineering simulation, especially cloud-native simulation, enables engineers to create more efficient and reliable impellers.

Basic Principles of Impeller Design

Impeller design uses fundamental fluid dynamics and energy transfer principles to function effectively. The primary function of an impeller is to convert mechanical energy from a motor into kinetic energy in the fluid. This process is governed by several fundamental principles and equations.

Bernoulli’s Equation

One of the foundational equations in fluid dynamics is Bernoulli’s equation, which describes energy conservation in a flowing fluid. It states that the total mechanical energy of the fluid remains constant along a streamline. The equation is given by:

$$ P = \frac{1}{2}\rho v^2 + \rho gh = constant $$

• \(P\) is the static pressure.
• \(\rho\) is the fluid density.
• \(v\) is the fluid velocity.
• \(g\) is the acceleration due to gravity.
• \(h\) is the height above a reference point.

Euler’s Turbomachinery Equation

Another critical principle is Euler’s turbomachinery equation, which relates the change in fluid energy to the impeller’s geometry and rotational speed. It is given by:

$$ \Delta H = \frac{U_2 V_{u2} – U_1 V_{u1}}{g} $$

• \(\Delta H\) is the head increase imparted to the fluid.
• \(U\) is the tangential velocity of the impeller.
• \(V_u\) is the tangential component of the absolute velocity of the fluid at the inlet (1) and outlet (2) of the impeller.

This equation is essential for determining the work done by the impeller on the fluid and is used to calculate the pressure increase provided by the impeller.

Key Design Parameters

The design of the impeller itself involves several key geometric parameters that influence its performance. These include:

• Impeller Diameter: The impeller diameter impacts both the head and flow rate. Larger diameters increase head and flow but also raise energy consumption. The relationship is approximated by the affinity law: \(H \propto D^2\)
• Blade Angle: Blade angles at the inlet (\(\beta_1)\) and outlet (\(\beta_2)\) are crucial for smooth fluid entry and exit, minimizing flow separation and turbulence. Optimized angles enhance energy transfer efficiency.
• Number of Blades: More blades reduce fluid slip and improve efficiency but increase manufacturing complexity. The optimal number balances efficiency and practical considerations.
• Blade Shape and Curvature: Curved blades guide fluid better than straight ones, reducing turbulence and energy losses. The blade shape is tailored to specific applications, such as radial, mixed-flow, or axial-flow impellers.
• Impeller Width: Impeller width affects flow rate and efficiency. Wider impellers handle larger flow rates but may increase friction losses. Narrower impellers are more efficient but support lower flow rates.
• Material Selection: Material choice impacts durability and resistance to wear and corrosion. Common materials include stainless steel, cast iron, and various alloys, selected based on operating conditions.
• Surface Finish: A smooth surface finish on blades and shrouds reduces friction and turbulence, enhancing hydraulic efficiency. Precision casting and surface coatings can improve the surface finish.

Types of Impellers

Table 1 below shows the types of impellers used in turbomachinery equipment (pumps or turbines).

Role of Engineering Simulation in Impeller Design

Simulating and evaluating a pump impeller early in the design process is crucial for determining the optimal design. However, traditional on-premises simulation tools are often costly and have steep learning curves.

Cloud-native simulation solutions offer significant advantages over traditional on-premises simulations. They provide scalable computing power, enabling engineers to run large-scale and complex simulations without local hardware limitations. Tools like SimScale eliminate these barriers by leveraging the power of the cloud.

With SimScale, engineers can:

• Optimize impeller designs faster by running several simulations simultaneously
• Use FEA and thermal analysis to test the stress and strain applied to pump impellers
• Get started quickly on an easy-to-use interface without extensive training
• Access cost-effective solutions and faster processing times

Solving Cavitation Problems in Pumps

Cavitation is the formation of vapor bubbles inside a liquid with low pressure and high flow velocity. It is the leading cause of performance deterioration in pumps and turbines, significantly affecting impellers.

SimScale offers advanced simulation tools that allow engineers to model and analyze cavitation effects in pumps and turbomachinery. Using SimScale, engineers can:

• Conduct Comprehensive Analyses: Utilize computational fluid dynamics (CFD) , structural (FEA) , and thermal analyses through automated workflows and intuitive interfaces.
• Model Cavitation Phenomena: Understand the impact of cavitation on performance by simulating cavitating flow and studying parameters like the net positive suction head required (NPSHR), cavitation number, and inlet sizing.
• Optimize Pump Efficiency: Use the Subsonic CFD solver to study and optimize pressure drop, fluid flow patterns, and cavitation effects. The solver’s robust meshing strategy ensures high-quality meshes and faster simulations.
• Import and Edit CAD Models: Easily import CAD models from various software and perform essential operations like flow volume extraction and defining rotating zones.
• Visualize Results: Leverage advanced visualization tools to analyze flow behavior, pressure distribution, velocity vectors, and cavitation effects.

Tecnalia’s Use of Cloud-Native Simulation in UAS Development

Tecnalia is a leading global research firm and Spain’s largest applied research and technological development organization. It designs and develops new technologies for drones and unmanned aircraft systems (UAS) for inspection, transport, and maintenance tasks. Tecnalia collaborates with industrial partners, local administrations, and end-users to improve and optimize transportation vehicles and their integration into urban ecosystems. It aims to develop efficient, environmentally friendly solutions for future sustainable and intelligent urban mobility.

Using SimScale, the Tecnalia team conducted detailed CFD simulations to optimize the design early in the process for the aerial vehicle from the Concept Integrity demonstrator. Extensive CFD analysis ensured aerodynamic efficiency and stable performance in various conditions, saving significant time and cost compared to physical testing. Read more in Tecnalia’s success story with SimScale .

Optimize Impeller Design With SimScale Cloud-Native Simulation

Are you seeking faster innovation and higher impeller design efficiency? SimScale offers a robust solution for reducing the time and cost associated with design and prototyping while maximizing accuracy and enhancing decision-making.

For engineers and designers aiming to push the boundaries of impeller design, SimScale provides the flexibility to explore innovative turbomachinery modeling . With SimScale’s integration with CFturbo , users can boost their impeller design modeling, benefiting from a seamless workflow that allows for faster and more accurate design and simulations.

SimScale’s advanced simulation capabilities enable you to test and validate innovative concepts that might be impractical or too risky to prototype traditionally. Experience the power of cloud-native simulation with SimScale. Sign up below and start simulating today.

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A Reds-A's trade to solve David Bell's Alexis Diaz problem

By zach pressnell | may 15, 2024.

To put it simply, the Cincinnati Reds have an Alexis Diaz problem. During the first half of 2023, their All-Star closer was lights out. Across his first 45 appearances of 2023, he held an ERA of 1.85 and had slammed the door nearly every time he took the mound.

Since that day, Diaz has been horrendous for Cincinnati. Across the last 42 appearances, Diaz has suffered seven losses, has an ERA near six and has only recorded 16 saves. In 2024, Diaz's ERA is a 6.75 and he's blown two of his nine save attempts. His 7.4 BB/9 is a horrendous number for Reds fans to look at.

What better way to solve this problem than by adding one of the game's most dominant bullpen arms of 2024?

A Reds-A's trade that would solve the Alexis Diaz problem in Cincinnati

There isn't a more popular bullpen arm in the media than that of the Oakland A's Mason Miller. Miller has seen his name attached to teams like the Red Sox, Cubs, and Phillies. Now, the idea of moving him to the Reds can be pitched.

The asking price for Miller is higher than most teams may be willing to pay. But most teams don't have the prospect capital that Cincinnati does. Last season, the Reds held one of the top farm systems in the entire league, graduating names like Christian Encarnacion-Strand, Elly De La Cruz and Matt McLain.

Reds-A's trade for Mason Miller

To acquire a 25-year-old closer like Mason Miller, Cincinnati could put together a trade package of Lyon Richardson and Blake Dunn to send to Oakland. Both Dunn and Richardson are MLB ready prospects listed in the Reds top 15 prospects, according to MLB Pipeline.

Oakland may be more attracted to some of the Reds' younger, higher ceiling prospects like Carlos Jorge or Hector Rodriguez, but it's quite unlikely that Cincinnati's front office would part ways with those two for a reliever, no matter how dominant.

Cincinnati could absolutely use the services of an improving 25-year-old closer that's been as impressive as Miller has been. Adding him to the bullpen would take stress off the shoulders of Alexis Diaz, giving Cincinnati a trio of Fernando Cruz, Alexis Diaz and Mason Miller that could shut down games in the Queen City throughout 2024.

Oakland will likely look to capitalize on Miller's value being higher than it ever will be. Cincinnati needs to start working to turn their train wreck of a season around before it's too late.

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Mathematics > Analysis of PDEs

Title: the obstacle problem for linear scalar conservation laws with constant velocity.

Abstract: In this contribution, we present a novel approach for solving the obstacle problem for (linear) conservation laws. Usually, given a conservation law with an initial datum, the solution is uniquely determined. How to incorporate obstacles, i.e., inequality constraints on the solution so that the resulting solution is still "physically reasonable" and obeys the obstacle, is unclear. The proposed approach involves scaling down the velocity of the conservation law when the solution approaches the obstacle. We demonstrate that this leads to a reasonable solution and show that, when scaling down is performed in a discontinuous fashion, we still obtain a suitable velocity - and the solution satisfying a discontinuous conservation law. We illustrate the developed solution concept using numerical approximations.

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