Question

A train is traveling up a 5.2° incline at a speed of 3.09 m/s when the last car breaks free and begins to coast without friction.
(a) How long does it take for the last car to come to rest momentarily?
(b) How far did the last car travel before momentarily coming to rest?

Answer 1

Initial velocity, u = 3.09 m/s
Final velocity, v = 0 m/s
acceleration due to gravity, g = 9.8 m/s/s
Acceleration, a = -g sin(5.2°)
Distance before coming to rest, S

Time to come to rest = (v-u)/a

Using
v²-u² = 2as
S can be found
S=(v²-u²)/2a

Answer 2

thanks much! ^^

Answer 3

You're welcome!