# A bus left the city traveling an avg. of 60 m/h. An hour later a car left at 75 m/h. How long will it take the car to catch the bus?

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Let's apply the formula of average speed which is

`s= d/t`

where s - average speed , d - distance travelled and t - time of travel.

For the bus, let `s_1` be its average speed and t its time of travel.Then, express its distance travelled in terms of t. So,

`d_1 = s_1 t`

Substituting value of `s_1` yields:

`d_1=60t`

And for the car , let `s_2` be its average speed. Since the car left one hour after the bus left, its time of travel is one less than t (time of bus).In math form, it is express as t-1. So the equation of the distance travelled by the car in terms of t is:

`d_2=s_2(t-1)`

`d_2=75(t-1)`

To simplify, distribute 75 to t-1.

`d_2=75t-75`

To solve for he time of travel of the car, when it caught up with the bus, set the two distances equal to each other.

`d_1=d_2`

`60t=75t-75`

Then. combine like terms. To do so, subtract both sides by 75t.

`60t-75t=75t-75t - 75`

`60t-75t=-75`

`-15t=-75`

And, divide both sides by -15 to isolate t.

`(-15t)/(-15)=(-75)/(-15)`

`t=5`

This means that the bus has already travelled for 5 hours when the car caught up with it.

Since the car left one hour after, then its time of travel is:

`t - 1 = 5-1 = 4`

**Hence, it took the car 4 hours to catch up with the bus.**