Question

Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.How far must the faster car travel before it has a 15-min lead on the slower car

Answer 1

The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.

Step-by-step explanation:

Given;

speed of the faster car, v₁ = 60 mi/h

speed of the slower car, v₂ = 55 mi/h

Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles

`(x)/(55) - (x)/(60) = (15)/(60)`

Note: divide 15 mins by 60 to convert to hours for consistency in the units.

`(x)/(55) - (x)/(60) = (15)/(60)\\\\multiple \ through \ by \ 660\\\\12x - 11x = 165\\\\x = 165 \ miles`

Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.