d2 = d1+29
Vo2*t + 0.5a*t^2 = Vo1*t + 29 m
42t + 1.5t^2 = 28t + 29
1.5t^2 + 42t - 28t = 29
1,5t^2 + 14t = 29
1.5t^2 + 14t - 29 = 0
Use Quadratic Formula and get:
t = 1.745 s. To pass the train.
V = Vo + a*t = 42 + 3*1.745 = 47.24 m/s.
A train is moving parallel and adjacent to a highway with a constant speed of 28 m/s. Initially a car is 29 m behind the train, traveling in the same direction as the train at 42 m/s and accelerating at 3 m/s^2.
What is the speed of the car just as it passes the train?
1 answer
1 answer