• Insufficient information. *** (Constant) acceleration is calculated using the equation [ (Vfinal - Vinitial) = AT ] *** We should re-write that to solve for acceleration, [ A = (Vfinal - Vinitial) / T ] *** Let's imagine an unreal scenario. The fellow accelerates to 70mph instantly. T = 0. The fellow undergoes infinite acceleration. OK. What about T = 1 second? A = 70mph / 1sec * 1h/60m * 1m/60s * 1609meter/mile = 31.3 m/s/s or 3.19g (one g = 9.81 m/s/s). What about 10 seconds? 0.319g. *** but note that this assumes that everything is perfectly solid and rebounding and non-stretching - sort of like a billiard ball. However: in the real world, the bag is going to stretch and the person in the bag is going to compress and so the actual number of gs ought to be slightly less than the sterile, unrealistic mathematical calculations show.
    • Sweet Deer
      Say it is non-compressible metal box inside a non-stretchable metal box?
      Yeah, something along those lines would fit the equations much better, but my guess is that several more impossible properties would have to be assigned to the items involved in order for the scenario to match the equations EXACTLY. For example: everything occurring in a perfect vacuum; an unbreakable and zero-ductility hook on the train to "snag" your box, a train car with similar properties to your boxes, then there's the car frame and axles and wheels and bearings and tracks and track bed and so on and so on
  • Interesting question. I agree that there is insufficient information to complete this as an exercise. If you knew specifically the time to get the person's body up to speed, you could just divide the change in velocity by the time to affect that change and there you go. I think that the realistic part of the problem is that the bag doesn't behave as a rigid body, that is to say that the string that transfers from the hook to the train accelerates more quickly, and then the weight hanging from the bottom accelerates more slowly, since a) that is where the most mass is, so it'll swing toward the back of the train as the string gains speed, and b) the string has to transfer the force of the collision by gaining tension, and then transfer that tension to the bag, which experiences a stress that ultimately transfers to the person inside. Since I have no idea what the weight of the person is, nor the length of the string on the mailbag, nor the modulus of elasticity of the string, I really couldn't say, but, if you know these things, you could calculate it the long way around.

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