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How fast is a freight train travelling?

Physics train problem?

  • The engineer driving the Shinkansen (Japanese bullet train) from Tokyo to Kyoto is having a nice day until he rounds a bend and suddenly sees a slow-moving freight train on his track. Both trains are travelling west, with speeds vs and vf respectively. Of course, as soon as the Shinkansen engineer sees the freight train, he applies the brakes; at that instant, the distance between the two trains is D. Show that if a collision is to be avoided, the magnitude of the Shinkansen's acceleration must be at least a= (Vs-Vf)^2/2D

  • Answer:

    Suppose the obstruction were stationary then the standard equations of motion will be used. To avoid a crunch D ≤ ½Vs t t = Vs/a so D ≤ Vs² / 2a a ≥ Vs² / 2D But the obstruction is moving at Vf So we change the inertial frame by declaring that the freight train is stationary. The ground is moving backwards at Vf and the Shinkansen is moving at Vs - Vf so a ≥ (Vs - Vf)² / 2D If the acceleration is the minimum then the trains will touch at speed Vf Of course in real life the driver of the Shinkansen would be whistling like crazy and the freight driver would apply full power. It wouldn't make much difference but it would help.

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That would only make sense if both trains begin stopping at the same time. (Vs-Vf) = velocity at which Shinkansen train closes in on freight train. Kinematic equation: Vfinal^2 = Vinitial^2 + 2 * a * d Here: Vfinal = 0 Vinitial = (Vs-Vf) d = D So: 0^2 = (Vs-Vf)^2 + 2 * a * D Assume a is negative due to DECELERATION 0 = (Vs-Vf)^2 - 2 * a * D 2 * a * D = (Vs-Vf)^2 a = (Vs-Vf)^2/2D But...yea if the other train keeps moving this isn't the very least the magnitude of acceleration would need to be.

Jeremy B

Suppose the obstruction were stationary then the standard equations of motion will be used. To avoid a crunch D ≤ ½Vs t t = Vs/a so D ≤ Vs² / 2a a ≥ Vs² / 2D But the obstruction is moving at Vf So we change the inertial frame by declaring that the freight train is stationary. The ground is moving backwards at Vf and the Shinkansen is moving at Vs - Vf so a ≥ (Vs - Vf)² / 2D If the acceleration is the minimum then the trains will touch at speed Vf Of course in real life the driver of the Shinkansen would be whistling like crazy and the freight driver would apply full power. It wouldn't make much difference but it would help.

Pavouk

That would only make sense if both trains begin stopping at the same time. (Vs-Vf) = velocity at which Shinkansen train closes in on freight train. Kinematic equation: Vfinal^2 = Vinitial^2 + 2 * a * d Here: Vfinal = 0 Vinitial = (Vs-Vf) d = D So: 0^2 = (Vs-Vf)^2 + 2 * a * D Assume a is negative due to DECELERATION 0 = (Vs-Vf)^2 - 2 * a * D 2 * a * D = (Vs-Vf)^2 a = (Vs-Vf)^2/2D But...yea if the other train keeps moving this isn't the very least the magnitude of acceleration would need to be.

Jeremy B

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