# Physics

When a high-speed passenger train traveling at 124 km/h rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance D=666 m ahead. The locomotive is moving at 21 km/. The engineer of the high-speed train immediately applies the brakes.

A) What must be the magnitude of the resulting constant deceleration of a collision is to be just avoided?

This is what I did:

I changed 124 km/h to 34.4 m/s and changed 21 km/h to 5.83 m/s. I believe at this point I am supposed to use one of the kinematic equations, but I am not sure which one. I think delta x = 34.4 - 5.83 = 28.57m. That would give me a position and I don't think there is an acceleration/deceleration rate so any equation with an a can't be used? That leaves me with this equation: delta x = 1/2(Vo + V)t

B) Assume the engineer is at x=0 when, at t=0, he first spots the locomotive. Sketch x(t) curves for the locomotive and high-speed train for the cases in which a collision is just avoided and is not quite avoided.

I'm struggling with velocity time graphs and I am not sure about this, but an explanation would be nice.

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1. Equations for constant acceleration:
v = Vi + a t
x = Xo + Vi t + (1/2) a t^2

here
for the high speed train
starts at 0 meters if locomotive starts at 666 meters
acceleration = negative a
v = 34.4 - a t until v = 5.83
so
5.83 = 34.4 - a t
a t = 28.6
x = 0 + 34.4 t -.5 a t^2
for the locomotive, no acceleration
v = 5.83
x = 666 + 5.83 t + 0 = 666 + 5.83 t
same x if they are at point x when their speeds are the same
so
34.4 t - .5 a t^2 = 666 + 5.83 t

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2. oh, not even a quadratic because at=28.6
34.4 t - .5(28.6) t = 666 + 5.83 t

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3. When I solve the part in the second post I get 46.67 sec. When I solve the quadratic I get -1014.14 and 1071.28
I arranged it as -0.5at^2 + 28.57 - 666 = 0
Also, I guess the wording of the question with "magnitude of the resulting constant deceleration" made me think the answer should be acceleration in m/s^2 but this is not the case as the answer is time in seconds. Why is this?

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4. The units are wrong, it did ask for an acceleration.

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