## A truck traveling at an average rate of 45 miles per hour leaves a rest stop. Fifteen minutes later s car traveling at an average rate of 60

Question

A truck traveling at an average rate of 45 miles per hour leaves a rest stop. Fifteen minutes later s car traveling at an average rate of 60 miles per hour leaves the same rest stop traveling the same route. How long will it take for the car to catch up with the truck

0

1. We write the kinematics equations
a
vf = a * t + vo
rf = (1/2) * a * t ^ 2 + vo * t + ro
We have then
For the truck:
a = 0
vf = vo
rf = 45t
For the car:
a = 0
vf = vo
rf = 60 * (t-15)
Matching rf:
45t = 60 * (t-15)
Clearing t:
45t = 60t-900
60t-45t = 900
15t = 900
t = 900/15 = 60h
it will take for the car to catch up with the truck 60h

2. Let t =  the time (hours) that the truck travels.
Because the car starts 15 minutes (or 0.25 hours) after the truck, the car travels for (t – 0.25) hours.

When the car catches up to the truck, both would have traveled the same distance.
The distance traveled by truck at 45 mi/h is
(45 mi/h)*(t h) = 45t mi.
The distance traveled by car at 60 mi/h is
(60 mi/h)*(t-0.25 h) = 60t – 15 mi.

Equate the two distances.
60t – 15 = 45t
60t – 45t = 15
15t = 15
t = 1