math help!?

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.

1. Assuming that they start at the same point, how much sooner does the faster car arrive at a destination 13 mi away?

2. How far must the faster car travel before it has a 15-min lead on the slower car?

Thanks and please show me how you did it!

Answer 1

use the formula d = v * t.
t = d / v
you have the distance (13 mi) and speeds ( v ). use these to figure out the time it takes each one to reach the destination.

for part 2 use subscripts. for car 1
t1 = d1 / v1
for car 2
t2 = d2 / v2

the difference in arrival time is t2 - t1 if t2 is the faster car
t2 - t1 = d/v1 - d / v2
you want the faster car to arrive 15 minutes before the slower
15 minutes = 0.25 hours
so t2 - t1 = 0.25 hours
you have d = 13 mi, v1 = 55mi/h
plug thes into the equation and solve for v2