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

Answer:

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

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

$\frac{x}{55} - \frac{x}{60} = \frac{15}{60}$

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

$\frac{x}{55} - \frac{x}{60} = \frac{15}{60}\\\\multiple \ through \ by \ 660\\\\12x - 11x = 165\\\\x = 165 \ miles$

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