Posted by **Katy** on Wednesday, January 19, 2011 at 11:12am.

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

- physics -
**Henry**, Thursday, January 20, 2011 at 8:56pm
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.

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