How to calculate distance based on just difference of arrival times of objects with different velocities

Question

Assume two cars leave a city at the exact same time, on the same road, and one car is traveling 60 miles per hour and the other car is traveling 50 miles per hour.

If the slower car arrives at your location 1/2 hour after the first car, what is the distance from your location to the city of origin?  You do not know when the cars left the city, only when they arrive at your location.

Answer 1

This question is using the speeds that the vehicles travel to find the distance. So what you need to find first is the difference in distance that they travel over an hour. So, for every hour that the fast car travels, the slower car goes 10 miles fewer. We know that the faster car got there 30 minutes earlier, so we can calculate how far behind the slow car was by using the speed it travels, 50 mph. One half of 50 is 25, so the slow car was 25 miles behind when the fast car got to the city. The difference every hour is 10 minutes so this means that the fast car traveled 25 / 10 hours, or 2.5. Using this we can find the distance traveled. 2.5 hours times 60 mph is 150 miles. The cars traveled 150 miles from the start city to the new city. 

Answer 2

Recall that distance = rate x time or d = rt

 

For the first car, d = 60t

For the second car, d = 50(t+1/2) since it takes the 2nd car 1/2 hour longer to arrive at the same destination.

 

We can set these two equations equal to one another since they are both equations for d

60t = 50(t+.5)

 

And solve

60t = 50t + 25

10t = 25

t = 2.5

 

So it takes the first car 2.5 hours to reach the destination.

 

Use this to solve for d

d = rt

d = 60(2.5)  or d = 50(2.5+.5)

d = 150 miles or using the other equation d=50(3) = 150 miles.

 

Hope this helps.