In these lessons, we will learn to solve word problems involving distance, rate (speed) and time.
Related Pages
Rate Distance Time Word Problems
Distance Word Problems
Average Speed Problems
Distance problems are word problems that involve the distance an object will travel at a certain average rate for a given period of time.
The formula for distance problems is:
distance = rate × time or
d = r × t
Things to watch out for:
Make sure that you change the units when necessary. For example, if the rate is given in miles
per hour and the time is given in minutes then change the units appropriately.
It would be helpful to use a table to organize the information for distance problems. A table helps you to think about one number at a time instead being confused by the question.
The following diagrams give the steps to solve Distance-Rate-Time Problems. Scroll down the page for examples and solutions.
We will show you how to solve distance problems by the following examples:
Example:
A bus traveling at an average rate of 50 kilometers per hour made the trip to town in 6 hours.
If it had traveled at 45 kilometers per hour, how many more minutes would it have taken to make
the trip?
Solution:
Step 1: Set up a rtd table.
r | t | d | |
Case 1 | |||
Case 2 |
Step 2: Fill in the table with information given in the question.
A bus traveling at an average rate of 50 kilometers per hour made the trip to town in 6 hours. If it had traveled at 45 kilometers per hour, how many more minutes would it have taken to make the trip?
Let t = time to make the trip in Case 2.
r | t | d | |
Case 1 | 50 | 6 | |
Case 2 | 45 | t |
Step 3: Fill in the values for d using the formula d = rt
r | t | d | |
Case 1 | 50 | 6 | 50 × 6 = 300 |
Case 2 | 45 | t | 45t |
Step 4: Since the distances traveled in both cases are the same, we get the equation:
45t = 300
Isolate variable t
Step 5: Beware - the question asked for “how many more minutes would it have taken to make the trip”, so we need to deduct the original 6 hours taken.
Answer: The time taken would have been 40 minutes longer.
Example:
This motion problem (or distance rate time problem or uniform rate problem) involves traveling
in the same direction, solving for “how long” one moving object traveling until it meets up with
the second moving object.
It uses d = rt (distance equals rate times time).
Car 1 starts from point A and heads for point B at 60 mph. Fifteen minutes later, car 2 leaves the same point A and heads for point B at 75 mph. How long before car 2 overtakes car 1?
Example:
A bus and a car leave the same place and traveled in opposite directions. If the bus is traveling
at 50 mph and the car is traveling at 55 mph, in how many hours will they be 210 miles apart?
Solution:
Step 1: Set up a rtd table.
r | t | d | |
bus | |||
car |
Step 2: Fill in the table with information given in the question.
If the bus is traveling at 50 mph and the car is traveling at 55 mph, in how many hours will they be 210 miles apart?
Let t = time when they are 210 miles apart.
r | t | d | |
bus | 50 | t | |
car | 55 | t |
Step 3: Fill in the values for d using the formula d = rt
r | t | d | |
bus | 50 | t | 50t |
car | 55 | t | 55t |
Step 4: Since the total distance is 210, we get the equation:
50t + 55t = 210
105t = 210
Answer: They will be 210 miles apart in 2 hours.
This motion problem (or distance rate time problem or uniform rate problem) involves one object traveling in one direction and the other in the opposite direction, solving for “how long” (or the amount of time) two moving objects traveling until they are certain distance apart.
Example:
Two planes leave the same point at 8 AM. Plane 1 heads East at 600 mph and Plane 2 heads West at
450 mph. How long will they be 1400 miles apart? At what time will they be 1400 miles apart?
How far has each plane traveled?
Example:
John took a drive to town at an average rate of 40 mph. In the evening, he drove back at 30 mph.
If he spent a total of 7 hours traveling, what is the distance traveled by John?
Solution:
Step 1: Set up a rtd table.
r | t | d | |
Case 1 | |||
Case 2 |
Step 2: Fill in the table with information given in the question.
John took a drive to town at an average rate of 40 mph. In the evening, he drove back at 30 mph. If he spent a total of 7 hours traveling, what is the distance traveled by John?
Let t = time to travel to town.
7 – t = time to return from town.
r | t | d | |
Case 1 | 40 | t | |
Case 2 | 30 | 7 – t |
Step 3: Fill in the values for d using the formula d = rt
r | t | d | |
Case 1 | 40 | t | 40t |
Case 2 | 30 | 7 – t | 30(7 – t) |
Step 4: Since the distances traveled in both cases are the same, we get the equation:
40t = 30(7 – t)
Use distributive property
40t = 210 – 30t
Isolate variable t
40t + 30t = 210
70t = 210
Step 5: The distance traveled by John to town is
40t = 120
The distance traveled by John to go back is also 120
So, the total distance traveled by John is 240
Answer: The distance traveled by John is 240 miles.
Example:
Roy took 5 hours to complete a journey. For the first 2 hours, he traveled at an average speed
of 65 km/h. For the rest of the journey, he traveled at an average speed of 78 km/h. What was
the total distance of the journey?
There is another group of distance-time problems that involves the speed of the water current or the speed of wind affecting the speed of the vehicle. The following video shows an example of such a problem.
How to solve Wind Word Problems?
Example:
Into the headwind, the plane flew 2000 miles in 5 hours. With a tailwind, the return trip took 4
hours. Find the speed of the plane in still air and the speed of the wind.
How to find the speed of the current of a stream?
Example:
The speed of a boat in still water is 10 mph. It travels 24 miles upstream and 24 miles downstream
in 5 hours. What is the speed of the current?
How to solve Current Word Problems?
Example:
Traveling downstream, Elmo can go 6 km in 45 minutes. On the return trip, it takes him 1.5 hours.
What is the boat’s speed in still water and what is the rate of the current?
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