car:
v = 36 + 4 t
x = 35 + 36 t + 2 t^2
train
v = 22
x = 22 t
the time t and the position x are the same at passing
22 t = 35 + 36 t + 2 t^2
solve quadratic for t
go back and find v = 36+4t
A train is moving parallel and adjacent to a highway with a constant speed of 22 m/s. Ini- tially a car is 35 m behind the train, traveling in the same direction as the train at 36 m/s and accelerating at 4 m/s2.
What is the speed of the car just as it passes the train?
Answer in units of m/s
1 answer
1 answer