A train is traveling up a 4° incline at a speed of 3.30m/s..?
..when the last car breaks free and begins to coast without friction.
How long does it take for the last car to come to rest momentarily?
How far did the last car travel before momentarily coming to rest?
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Vertical component of acceleration of last car is g sin 4° = 0.684 m/s², to 3 significant figures.
So the car will take approximately 3.30/0.684 = 4.82 s to momentarily come to rest.
(Using equation for uniformly accelerated motion v = u + at -- see link below.)
During that time it will travel approximately ½ × 3.30 × 4.82 = 7.95 m.
(Using equation for uniformly accelerated motion s = ½(u + v)t.)