When a high-speed passenger train traveling at 44.72 m/s rounds a bend, the engi

Question

When a high-speed passenger train traveling at 44.72 m/s 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 = 676 m ahead (See Above Figure). The locomotive is moving at 8.056 m/s. The engineer of the high-speed train immediately applies the brakes.

Assuming the high-speed train is moving with an acceleration of (-0.9942 m/s2), answer the question below. Show all work and give explanation.

Find out the speed of the train when it is at the distance of x = ½D from the locomotive.

Explanation / Answer

consider the position of the train at the origin

Vf = final speed when the distance between train and locomotive is half

Vo = initial speed = 44.72 m/s

Xo = initial position of the train = 0

X = final position of train

a = acceleration = - 0.9942

t = time of travel

using the equation

Xf = Xo + Vo t + (0.5) a t2

Xf = 0 + 44.72 t + (0.5) (- 0.9942) t2 eq-1

for the locomotive :

X'o = initial position = D

X'f = final position

V'o = velocity of locomotive = 8.056 m/s

X'f = X'o + V'o t

X'f = D + 8.056 t eq-2

Given that

X'f - Xf = D/2

D + 8.056 t -  44.72 t - (0.5) (- 0.9942)  t2 = D/2

676 + 8.056 t -  44.72 t - (0.5) (- 0.9942)  t2 = 676/2

t = 10.8 sec

velocity after 10.8 sec for train is given as

Vf = Vo + a t

Vf = 44.72 + (-0.9942) (10.8)

Vf = 34 m/s

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