An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.50 m above the ground. The railcar has a mass of 38500 kg and is moving to the right at a constant speed of 8.70 m/s on a frictionless rail. The automobile them accelerates to the left, leaving the railcar at a speed of 22.0 m/s with respect to the ground. When the automobile lands, what is the distance D between it and the left end of the railcar?

asked by Khalid

9 years ago

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1 answer

t to fall is sqrt(2x/g) = .56
distance car moves = .55*22 = 12.17
rail car moves .55*(8.7)= 4.78
tot = 12.17+4.78

answered by Chanz

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Question

An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.50 m above the ground. The railcar has a mass of 38500 kg and is moving to the right at a constant speed of 8.70 m/s on a frictionless rail. The automobile them accelerates to the left, leaving the railcar at a speed of 22.0 m/s with respect to the ground. When the automobile lands, what is the distance D between it and the left end of the railcar?

1 answer

t to fall is sqrt(2x/g) = .56
distance car moves = .55*22 = 12.17
rail car moves .55*(8.7)= 4.78
tot = 12.17+4.78

Chanz · Answer

t to fall is sqrt(2x/g) = .56 distance car moves = .55*22 = 12.17 rail car moves .55*(8.7)= 4.78 tot = 12.17+4.78