A train is traveling down a straight track at 34 m/s when the engineer applies t

Question

A train is traveling down a straight track at 34 m/s when the engineer applies the brakes, resulting in an acceleration of -1.0 m/s2 as long as the train is in motion. How far does the train move during a 23 s time interval starting at the instant the brakes are applied?

I arrived at 60.5 and was told my answer is off by more than 10%.


A certain freely falling object requires 1.20 s to travel the last 35.0 m before it hits the ground. From what height above the ground did it fall?

I arrived at 6.3 and was told my answer is off by a multiple of 10

Explanation / Answer

A train is traveling down a straight track at 34 m/s when the engineer applies the brakes, resulting in an acceleration of -1.0 m/s2 as long as the train is in motion. How far does the train move during a 23 s time interval starting at the instant the brakes are applied? Vo = 34 a = -1 t=23 Vf=Vo+at Vf=34-1(23) Vf=11 Vf^2=Vo^2+2ax solve for x x=(Vf^2-Vo^2)/2a plug in our numbers x=517.5 A certain freely falling object requires 1.20 s to travel the last 35.0 m before it hits the ground. From what height above the ground did it fall? x=35 t=1.2 a=-9.81 This Vo is the speed right at 35 m above the ground x=Vot+(1/2)at^2 vo=(x-(1/2)at^2)/t Vo=35.05 m/s now lets find the distance from which it fell. When it is at its peak height, the velocity in the y direction is zero. I assume that its in free fall and that all the velocity is in the y direction. so Vo=0 and Vf=35.05 so we know how far it went. a=-9.81 Vf^2=Vo^2+2ax Vo = 0 solve for x x=(Vf^2)/2a x=-62.62 its negative because its moving downwards, we wont count that because we just want its max height. so we add the last 35m to this. 35+62.62 = x x = 97.62m

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