Question

A train which is traveling at 73mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its decreased velocity is observed to be 56mi/hr as it passes a point 1/2mi beyond A. A car moving at 45mi/hr passes point B at the same instant that the train reaches point A. In an unwise effort to beat the train to the crossing, the driver "steps on the gas." Calculate the constant acceleration a that the car must have in order to beat the train to the crossing by 3.9sec and find the velocity v of the car as it reaches the crossing.

Answer 1

First to find deceleration of the train we use

v²= u²+ 2as

56²= 73²+ 2(0.5)a

a= -2193mi/hr²

Then we find time in which the train does the intersection

Using

S= ut+ 1/2 at²

1= 73t-1/2(1293)t²

t =68.5s

But since the train is to intersect in 3.9s the time will be the difference which is

65.68s

So finding acceleration

S= ut + 1/2at²

1.3mi= 45/3600mi/s(65.58s)+ 1/2a(65.5)²

So a= 1.179ft/s²

To find velocity we use

V= u + at

= 45/3600mi/s + (-2.33E-4mis²)(65.58s)

V= 0.0271mi/s

= 97.6ft/s