A train is moving parallel and adjacent to a highway with a constant speed of 27 m/s. Ini- tially a car is 60 m behind the train, traveling in the same direction as the train at 41 m/s and accelerating at 5 m/s2.

Question

A train is moving parallel and adjacent to a highway with a constant speed of 27 m/s. Ini- tially a car is 60 m behind the train, traveling in the same direction as the train at 41 m/s and accelerating at 5 m/s2.
What is the speed of the car just as it passes the train?
Answer in units of m/s

Henry · Answer

Dc = Vo*t + 0.5a*t^2 = 27t+60
41t + 0.5*5*t^2 = 27t + 60
41t -27t + 2.5t^2 = 60
14t + 2.5t^2 = 60
2.5t^2 + 14t - 60 = 0
Use Quadratic formula.
t = 2.84 s. To catch up.

Vc = Vo + a*t = 41 + 5*2.84 = 55.2 m/s.