   Chapter 2, Problem 10P

Chapter
Section
Textbook Problem

Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 70 mi/h. (a) Assuming they start at the same point, how much sooner does the faster car arrive at a destination 10 mi away? (b) How far must the faster car travel before it has a 15-min lead on the slower car?

(a)

To determine
The time taken for the faster car to reach the destination sooner than the slower car.

Explanation

Given info: The distance covered by both the cars is 10mi , the speed of the faster car is 70mi/h , and the speed of the slower car is 55mi/h .

Explanation:

The formula used to calculate the difference in time between the cars to reach the destination is,

Δt=Δxv1Δxv2

Here,

Δt is the difference in time between the cars to reach the destination.

Δx is the distance covered by the cars.

v1 is the speed of the slower car.

v2 is the speed of the faster car.

Substitute 10mi for Δx , 55mi/h for v1 and 70mi/h for v2 to find Δt

(b)

To determine
The distance the faster car travelled when the car has a 15.0min lead over the slower car.

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