Question

Two cars travel westward along a straight highway, one at a constant velocity of 85 km/h, and the other at a constant velocity of 115 km/h.

a. assuming that both cars start at the same point, how much sooner does the faster car arrive at a destination 16 km away?
b. how far must the cars travel for the faster car to arrive 15 min before the slower car?

Answer 1

To solve letter a:

d1 = 85t1 = 16 km,
85t1 = 16,
t1 = 16 / 85 = 0.1882 h = 11.29 min.

d2 = 115t2 = 16 km,
115t2 = 16,
t2 = 16 / 115 = 0.139 h = 8.35 min.

t1 - t2 = 11.29 - 8.35 = 2.94 min.
Car #2 arrives 2.94 minutes sooner.

To solve letter b:

15 min = 1/4 h = 0.25 h.
d1 = d2,
115t = 85(t + 0.25),
115t = 85t + 21.25,
115t - 85t = 21.25,
30t = 21.25,
t = 21.25 / 30 = 0.71 h,

d = 115 * 0.71 = 81.65 km.