Help me with this question please it's due in a couple hours!!!!!?

Question

A high speed train traveling at 161 km/hr rounds a bend and is shocked to see that a locomotive has entered the tracks going the same direction .676 km ahead of it. The locomotive is traveling at 29.0 km/hr. The engineer of the high speed train applies the brakes. What must be the magnitude of the constant deceleration of the train so that a collision is just avoided?

Some equations that may help:

Final Velocity = Original Velocity + acceleration(time)

position = (1/2)acceleration(time squared) + original velocity(time)

Final velocity squared = original velocity squared + 2(acceleration)(position)

or

Vf = Vo + at

x = (1/2)at^2 + Vot

Vf^2 = Vo^2 + 2ax

Answer 1

You can use the last equation if you view things from the frame of reference of the locomotive. In that frame, the locomotive is at rest, and initially the high speed train is traveling toward the locomotive at (161-29) = 132 km/hr. We want Vf = 0, and we have x = 0.676. So

0 = 132² + 2a(0.676)

-132² = 1.352a

-132²/1.352 = a

-1.29×10^4 = a

The units are km/hr², and the negative sign is correct because the train is decelerating. Since you are asked the magnitude, you would drop the minus sign.

Divide the result for a by 3600² to get units of km/sec², and then multiply that by 1000 if you want acceleration in meters/sec²

You could also solve this from the frame of reference of the ground:

From this frame of reference, the first equation would have Vf = 29, and Vo=161; the second equation would have position = 0.676 + 29t and, again, Vo = 161 (both equations referring, of course, to the high speed train). The solution to these two equations would be the acceleration a, and the time t at which the high speed train would just touch (at zero velocity) the locomotive.