information in the problem.
[latex]r=12\text{mph}[/latex]
[latex]t=3 \Large\frac{1}{2}\normalsize\text{hours}[/latex]
Write the appropriate formula for the situation.
Substitute in the given information.
[latex]d=12\cdot 3\Large\frac{1}{2}[/latex]
In the following video we provide another example of how to solve for distance given rate and time.
Rey is planning to drive from his house in San Diego to visit his grandmother in Sacramento, a distance of [latex]520[/latex] miles. If he can drive at a steady rate of [latex]65[/latex] miles per hour, how many hours will the trip take?
Show Solution
Step 1. the problem. Summarize the information in the problem. | [latex]d=520[/latex] miles [latex]r=65[/latex] mph [latex]t=?[/latex] |
Step 2. what you are looking for. | how many hours (time) |
Step 3. Choose a variable to represent it. | let [latex]t[/latex] = time |
Step 4. Write the appropriate formula. Substitute in the given information. | [latex]d=rt[/latex] [latex]520=65t[/latex] |
Step 5. the equation. | [latex]t=8[/latex] |
Step 6. Substitute the numbers into the formula and make sure [latex]d=rt[/latex] [latex]520\stackrel{?}{=}65\cdot 8[/latex] [latex]520=520\quad\checkmark [/latex] | |
Step 7. the question with a complete sentence. We know the units of time will be hours because | Rey’s trip will take [latex]8[/latex] hours. |
In the following video we show another example of how to find rate given distance and time.
Calcworkshop
// Last Updated: January 20, 2020 - Watch Video //
An object is said to be in uniform motion when it moves without changing its speed.
Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)
All this means is that we can find the distance an object travels as long as we know the object is moving at a constant (fixed) speed or pace or at an average rate or speed.
In fact, whenever we encounter this type of scenario we can utilize a very powerful formula:
Distance-Rate-Time Example
The type of questions we will be investigating in this lesson involve:
Solving Distance Rate Time Example
And all three of these questions can be answered using the Distance formula listed above.
Yes, it is simple, but there is a warning.
Units of measure matter!
As Purple Math so accurately points out, we must always check our units to ensure they agree with one another.
Many questions will try to trick you by using miss-matched units, so always check and if necessary, convert to the correct unit before simplifying.
Together we will look at five examples in detail for how to solve Distance-Rate-Time word problems.
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Explanation More Examples
"Distance" word problems, often also called "uniform rate" problems, involve something travelling at some fixed and steady ("uniform") pace ("rate", "velocity", or "speed"), or else you are told to regard to object as moving at some average speed.
Content Continues Below
Distance Word Problems
Whenever you read a problem that involves "how fast", "how far", or "for how long", you should think of the distance equation, d = rt , where d stands for distance, r stands for the (constant or average) rate of speed, and t stands for time.
Make sure that the units for time and distance agree with the units for the rate. For instance, if they give you a rate of feet per second, then your time must be in seconds and your distance must be in feet. Sometimes they try to trick you by using mis-matched units, and you have to catch this and convert to the correct units.
In case you're wondering, this type of exercise requires that the rate be fixed and steady (that is, unchanging) for the d = rt formula to work. The only way you can deal with a speed that might be changing over time is to take the average speed over the time or distance in question. Working directly with changing speeds will be something you'll encounter in calculus, as it requires calculus-based (or more advanced) methods.
A fixed-speed exercise is one in which the car, say, is always going exactly sixty miles an hour; in three hours, the car, on cruise-control, will have gone 180 miles. An average-speed exercise is one in which the car, say, averaged forty miles an hour, but this average includes the different speeds related to stop lights, highways, and back roads; in three hours the car went 120 miles, though the car's speed was not constant. Most of the exercises you'll see will be fixed-speed exercises, but obviously they're not very "real world". It's a simplification they do in order to make the situation feasible using only algebraic methods.
There is a method for setting up and solving these exercises that I first encountered well after I'd actually been doing them while taking a class as an undergraduate. But, as soon as I was introduced to the method, I switched over, because it is *so* way easier.
First I set up a grid, with the columns being labelled with the variables from the "distance" formula, and the rows being labelled with the "parts" involved:
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---|---|---|---|
first part |
|
|
|
second part |
|
|
|
total |
|
|
|
In the first part, the plane covered some distance. I don't know how much, so I'll need a variable to stand for this unknown value. I'll use the variable they give me in the distance equation:
They gave me the speed, or rate, for this part, so I'll add this to my table:
|
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---|---|---|---|
first part |
| 105 |
|
second part |
|
|
|
total |
|
|
|
The plane flew for some amount of time during this first part, but I don't know how long that was. So I need a variable to stand for this unknown value; I'll use the variable from the distance equation:
For the second part, the plane travelled the rest of the total distance. I don't know the exact distance that was flown during this second part, but I do know that it was "however much was left of the 555 miles, after the first d miles were flown in the first part. "How much was left after [some amount] was taken out" is expressed with subtraction: I take the amount that has been taken care of already, and subtract this from whatever was the total. Adding this to my table, I get:
|
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|
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---|---|---|---|
first part |
| 105 |
|
second part | 555 − |
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|
total |
|
|
|
They've given me the speed, or rate, for the second part, and I can use the same "How much is left?" construction for whatever was the time for this second part. So now my table looks like this:
|
|
|
|
---|---|---|---|
first part |
| 105 |
|
second part | 555 − | 115 | 5 − |
total |
|
|
|
For the "total" row, I add down (or take info from the exercise statement):
|
|
|
|
---|---|---|---|
first part |
| 105 |
|
second part | 555 − | 115 | 5 − |
total | 555 | --- | 5 |
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Why did I not add down in the "rate" column? Because I cannot add rates! In this exercise, adding the rates would have said that the average rate for the entire trip was 105 + 115 = 220 miles per hour. But obviously this makes no sense.
The genius of this table-based method of set-up is that I can now create equations from the rows and columns. In this exercise, there is more than one way to proceed. I'll work with the "distance" equation to create expressions for the distances covered in each part.
Multiplying across, the first row tells me that the distance covered in the first part of the flight was:
1st part distance: 105 t
Again multiplying across, the second row tells me that the distance covered in the second part of the flight was:
2nd part distance: 115(5 − t )
I can add these two partial-distance expressions, and set them equal to the known total distance:
105 t + 115(5 − t ) = 555
This is an equation in one variable, which I can solve:
105 t + 115(5 − t ) = 555 105 t + 575 − 115 t = 555 575 − 10 t = 555 20 = 10 t 2 = t
Looking back at my table, I see the I had defined t to be the time that the plane spent in the air on the first part of its journey. Looking back at the original exercise, I see that they want to know the times that the plane spent at each of the two speeds.
I now have the time for the first part of the flight; the time was two hours. The exercise said that the entire trip was five hours, so the second part must have taken three hours (found by subtracting the first-part time from the total time). They haven't asked for the partial distances, so I now have all the information I need; no further computations are necessary. My answer is:
first part: 2 hours second part: 3 hours
When I was setting up my equation, I mentioned that there was more than one way to proceed. What was the other way? I could have used the table to create an expression for each of the two partial times, added, set the result equal to the given total, and solved for the variable d . Since the distance equation is d = rt , then the expressions for the partial times would be created by solving the equation for t = . My work would have looked like this:
first part: d /105
second part: (555 − d )/115
adding: d /105 + (555 − d )/115 = 5 23 d + 11,655 − 21 d = 12,075 2 d = 420 d = 210
Looking back at my table, I would have seen that this gives me the distance covered in the first half of the flight. Looking back at the exercise, I would have seeing that they are wanting times, not distances. So I would have back-solved for the time for the first part, and then done the subtraction to find the time for the second part. My work would have had more steps, but my answer would have been the same.
There are three things that I hope you take from the above example:
(My value for the distance, found above, is correct, but was not what they'd asked for.) But even more important to understand is this:
NEVER TRY TO ADD RATES! Think about it: If you drive 20 mph on one street, and 40 mph on another street, does that mean you averaged 60 mph? Of course not.
Can I even average the rates? If I drove at 20 mph for one hour, and then drove 60 mph for two hours, then I would have travelled 140 miles in three hours, or a little under 47 mph. But 47 is not the average of 20 and 60 .
As you can see, the actual math involved in solving this type of exercise is often quite simple. It's the set-up that's the hard part. So what follows are some more examples, but with just the set-up displayed. Try your hand at solving, and click on the links to get pop-ups from which to check your equations and solutions.
I will start in the usual way, by setting up my table:
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driving |
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flying |
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total |
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I have labelled my rows so it's clear how they relate to the exercise. Now I need to fill in the rows. As before, I don't know the distance or the time for the part where the executive was driving, so I'll use variables for these unknowns, along with the given rate.
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driving |
| 30 |
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flying |
|
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total |
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|
For the flying portion of the trip, I'll use the "how much is left" construction, along with the given rate, to fill in my second row. I'll also fill in the totals.
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---|---|---|---|
driving |
| 30 |
| flying | 150 − | 60 | 3 − |
total | 150 | --- | 3 |
The first row gives me the equation d = 30 t . The second row is messier, giving me the equation:
150 − d = 60(3 − t )
There are various ways I can go from here; I think I'll solve this second equation for the variable d , and then set the results equal to each other.
150 − d = 60(3 − t ) 150 − 60(3 − t ) = d
Setting equal these two expressions for d , I get:
30 t = 150 − 60(3 − t )
Solve for t ; interpret the value; state the final answer.
Both vehicles travelled for the same amount of time.
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---|---|---|---|
car |
|
| 2 |
bus |
|
| 2 |
total |
| --- |
|
The car's values are expressed in terms of the bus' values, so I'll use variables for the bus' unknowns, and then define the car in terms of the bus' variables. This gives me:
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|
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---|---|---|---|
car | + 20 | 2 − 30 | 2 |
bus |
|
| 2 |
total | --- | --- | --- |
(As it turns out, I won't need the "total" row this time.) The car's row gives me:
d + 20 = 2(2 r − 30)
This is not terribly helpful. The second row gives me:
I'll use the second equation to simplify the first equation by substituting " 2 r " from the second equation in for the " d " in the first equation. Then I'll solve the equation for the value of " r ". Finally, I'll need to interpret this value within the context of the exercise, and then I'll state the final answer.
(Remember that the expression for the car's speed, from the table, was 2 r − 30 , so all you need to do is find the numerical value of this expression. Just evaluate; don't try to solve — again — for the value of r .)
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Understanding motion and work problems, especially those involving the formula \(d=rt\) (where \(d\) stands for distance, \(r\) for rate, and \(t\) for time), can be simplified through a step-by-step guide:
Step 1: understanding the formula.
By following these steps and practicing regularly, understanding motion and work problems, especially those related to the distance formula \(d=rt\), becomes more manageable and intuitive.
Example 1 :
A car travels at a speed of \(50\) miles per hour. How far will it travel in \(3\) hours?
\(Distance = Rate × Time = 50 \ \frac{miles}{hour} × 3 \ hours = 150 \ miles\).
Example 2 :
A cyclist covers a distance of \(90\) kilometers in \(3\) hours. What is the cyclist’s average speed?
\(Speed = Distance ÷ Time = 90 \ km ÷ 3 \ hours = 30 \frac{km}{hour}\).
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Before you can use the distance, rate, and time formula, D=RT, you need to make sure that your units for the distance and time are the same units as your rate. If they aren't, you'll need to change them so you're working with the same units.
Algebra. Solve for r d=rt. d = rt d = r t. Rewrite the equation as rt = d r t = d. rt = d r t = d. Divide each term in rt = d r t = d by t t and simplify. Tap for more steps... r = d t r = d t. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just ...
D. Russell. Print the PDF: Distance, Rate, and Time Worksheet No. 2. If students are struggling, explain that to solve these problems, they will apply the formula that solves distance, rate, and time, which is distance = rate x time. It is abbreviated as: d = rt The formula can also be rearranged as: r = d/t or t = d/r
Rate is distance per time, so its units could be mph, meters per second, or inches per year. Now you can solve the system of equations: 50t = 100 (t - 2) (Multiply both values inside the parentheses by 100.) 50t = 100t - 200. 200 = 50t (Divide 200 by 50 to solve for t.) t = 4. Substitute t = 4 into train No. 1.
DISTANCE = RATE X TIME. Explore the formula d = rt by starting with unit conversion problems. Mathletes will solve for distance, rate and time by paying attention to the units given in the problem and using the appropriate equivalent version of the formula: d = rt, r = d / t or t = d / r. Download Mathlete handout. Download coach version with ...
30. t. 30 × t. The total distance you and your wife travel is 45t + 30t = 75t and this must be equal to 225. 75 × t = 225. Since 75 times 3 = 225, t = 3. Therefore, you will be 225 miles apart after 3 hours. The above are great examples of distance rate time problems. Try to do some now in your math textbook.
Distance, Rate, and Time. For an object moving at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula. d= rt d = r t. where d = d = distance, r = r = rate, and t= t = time. Notice that the units we used above for the rate were miles per hour, which we can write as a ratio miles hour m i l ...
In fact, whenever we encounter this type of scenario we can utilize a very powerful formula: d = rt. Distance-Rate-Time Example. Where: d represents distance. r is an object's rate (speed) t is the time the object takes to travel. The type of questions we will be investigating in this lesson involve:
Distance - Rate - Time Word Problems Date_____ Period____ 1) An aircraft carrier made a trip to Guam and back. The trip there took three hours and the trip back took four hours. It averaged 6 km/h on the return trip. Find the average speed of the trip there. 2) A passenger plane made a trip to Las Vegas and back.
Solve problems from Pre Algebra to Calculus step-by-step step-by-step. d=rt , for r. en. Related Symbolab blog posts. Practice Makes Perfect. Learning math takes practice, lots of practice. Just like running, it takes practice and dedication. If you want... Enter a problem. Cooking Calculators.
In 3 hours, they are 81 miles apart. Find the rate of each cyclist. Rate-Time-Distance Problem 2. Solve this word problem using uniform motion rt = d formula. Example: A jogger started running at an average speed of 6 mph. Half an hour later, another runner started running after him starting from the same place at an average speed of 7 mph.
2nd part distance: 115 (5 − t) I can add these two partial-distance expressions, and set them equal to the known total distance: 105 t + 115 (5 − t) = 555. This is an equation in one variable, which I can solve: 105 t + 115 (5 − t) = 555. 105 t + 575 − 115 t = 555. 575 − 10 t = 555. 20 = 10 t.
How to use the formula for distance, rate (speed), and time to solve problems. This video lesson is an introduction to the formula for distance, constant speed, and time: d = vt (or d = rt if you use r as rate instead of v as velocity). First I explain an easy way to memorize this formula in the form v = d/t, comparing it to the common way to express the speed of cars, such as 55 miles per hour.
Step 1: Determine what information given in the word problem represents d, distance, r, rate, and t, time. Step 2: Plug d, r, and t into the equation d = r × t . Step 3: Solve for the unknown ...
Understanding motion and work problems, especially those involving the formula \(d=rt\) (where \(d\) stands for distance, \(r\) for rate, and \(t\) for time), can be simplified through a step-by-step guide: Effortless Math. X + eBooks ... Practice Problems: Regularly solve different types of motion problems to get a better grasp of the concept.
Differentiation. dxd (x − 5)(3x2 − 2) Integration. ∫ 01 xe−x2dx. Limits. x→−3lim x2 + 2x − 3x2 − 9. Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more.
Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. ... Enter a problem... Algebra Examples. Popular Problems. Algebra. Solve for r d=rt for r. for . Step 1. Rewrite the equation as . Step 2. Divide each term in by and simplify ...
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Algebra questions and answers. Practice Problems 1. Solve d rt for r 2. Solve P=144p for p 3. Solve R.cs for C 4. Solve P-a+b+c for b 5. Solve T m-n for n 6. Solve A for b 7, Solve V = lvvh for w 8. Solve m=m for y2 9.
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Integration. ∫ 01 xe−x2dx. Limits. x→−3lim x2 + 2x − 3x2 − 9. Online math solver with free step by step solutions to algebra, calculus, and other math problems. Get help on the web or with our math app.
Solve for t D=rt. D = rt D = r t. Rewrite the equation as rt = D r t = D. rt = D r t = D. Divide each term in rt = D r t = D by r r and simplify. Tap for more steps... t = D r t = D r. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a ...
Polynomial. In mathematics, a polynomial is a mathematical expression consisting of indeterminates and coefficients, that involves only the operations of addition, subtraction, multiplication, and positive-integer powers of variables. An example of a polynomial of a single indeterminate x is x² − 4x + 7. An example with three indeterminates ...
Algebra. Solve for r rt=d. rt = d r t = d. Divide each term in rt = d r t = d by t t. rt t = d t r t t = d t. Simplify the left side. Tap for more steps... r = d t r = d t. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.
Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. ... Solve for r d=r/(1+rt) Step 1. Rewrite the equation as . Step 2. Find the LCD of the terms in the equation. Tap for more steps... Step 2.1. Finding the LCD of a list of ...