Atrain which is traveling at 70 mi/hr applies its brakes as it reaches point a and slows down with a constant deceleration. its decreased velocity is observed to be 52 mi/hr as it passes a point 1/2 mi beyond a. a car moving at 52 mi/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.
(a) calculate the constant acceleration a that the car must have in order to beat the train to the crossing by 4.3 sec.
(b) find the velocity v of the car as it reaches the crossing.

a) 0 mi/s^2

b) 52 mi/s

Explanation:

Assuming the crossing is 1/2 mile past point A and that point B is near point A (it isn't clear in the problem)

The train was running at 70 mi/h at point A and with constant deceleration reachesn the crossing 1/2 mile away with a speed of 52 mi/h

The equation for position under constant acceleration is:

X(t) = X0 + V0 * t + 1/2 * a * t^2

I set my reference system so that the train passes point A at t=0 and point A is X = 0, so X0 = 0.

Also the equation for speed under constant acceleration is:

V(t) = V0 + a * t

Replacing

52 = 70 + a * t

Rearranging

a * t = 52 - 70

a = -18/t

I can then calculate the time it will take it to reach the crossing

1/2 * a * t^2 + V0 * t  - X(t) = 0

Replacing

1/2 (-18/t) * t^ + 70 * t - 1/2 = 0

-9 * t + 70 * t = 1/2

61 * t = 1/2

t = (1/2)/61 = 0.0082 h = 29.5 s

And the acceleration is:

a = -18/0.0082 = -2195 mi/(h^2)

To beath the train the car must reach the crossing in 29.5 - 4.3 = 25.2 s

X(t) = X0 + V0 * t + 1/2 * a * t^2

52 mi/h = 0.0144 mi/s

1/2 = 0 + 0.0144 * 25.2 + 1/2 * a * 25.2^2

1/2 = 0.363 + 317.5 * a

317.5 * a = 0.5 - 0.363

a = 0.137/317.5 = 0.00043 mi/s^2 (its almost zero)

The car should remain at about constant speed.

It will be running at the same speed.

answer:

an example of solubility is salt or sugar dissolving in water.

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