45.0k views
4 votes
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

User PSyToR
by
8.0k points

1 Answer

3 votes

Answer:

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.

User Joshua Howard
by
8.1k points

No related questions found