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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
[tex]\frac{x}{55} - \frac{x}{60} = \frac{15}{60}[/tex]
Note: divide 15 mins by 60 to convert to hours for consistency in the units.
[tex]\frac{x}{55} - \frac{x}{60} = \frac{15}{60}\\\\multiple \ through \ by \ 660\\\\12x - 11x = 165\\\\x = 165 \ miles[/tex]
Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.