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Time Speed and Distance- Bus and A car- Part 1

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by [email protected] » Sun May 11, 2014 11:33 pm
Hi s91arvindh,

This questions requires that you use the Distance Formula and keep track of 2 moving vehicles.

The bus is moving 60mph and is in motion for 20 minutes....

D = R x T
D = 60mph x 1/3 hour

D = 20 miles

So we know the bus traveled 20 miles.

We also know that the bus started 5 miles behind a car, but ended 10 miles ahead of that car. So the bus "made up" 15 miles on the car, even though the bus traveled 20 miles. This means that the car had to have been moving; since the bus "made" up 15 miles instead of 20 miles, the car must have traveled 5 miles during that time.

D = R x T
5 miles = R x 1/3 hour

[spoiler]R = 15mph[/spoiler]

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by GMATGuruNY » Mon May 12, 2014 2:33 am
s91arvindh wrote:A bus moving at 60 mph is 5 miles behind a car. After 20 minutes, it is 10 miles ahead of the car.
what is the speed of the car .

15mph
The CATCH-UP rate is the DIFFERENCE between the two rates:
Catch-up rate = (faster rate) - (slower rate).

In 1/3 of an hour, the bus catches up a total of 15 miles (from 5 miles behind to 10 miles ahead).
Thus:
The catch-up rate = d/t = 15/(1/3) = 45 miles per hour.

Since 45 miles per hour is difference between the bus's rate and the car's rate, we get:
45 = 60 - (car's rate)
Car's rate = 15 miles per hour.
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