physics
posted by Katy .
A train is moving parallel and adjacent to a
highway with a constant speed of 22 m/s. Initially a car is 69 m behind the train, traveling
in the same direction as the train at 35 m/s
and accelerating at 3 m/s
2
.
What is the speed of the car just as it passes
the train

D1 = D2  69.
V*t = Vo + 0.5at^2 69,
22t = 35t + 0.5 * 3t^2  69,
22t = 35t + 1.5t^2  69,
22t  1.5t^2  35t = 69,
1.5t^2 13t + 69 = 0,
Multiply both sides by 1:
1.5t^2 + 13t  69 = 0.
Using Quadratic Formula,
t = (13 + sqrt(169 + 414)) / 3,
t = (13 + 24.1) / 3,
t = 3.72, and 12.37.
Use the positive value of t:
t = 3.72s.
V = Vo + at = 35 + 3 * 3.72 = 46.12m/s.