Long Train
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A 90 m long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/s?
A 90 m long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/s when it passes a railway worker who is standing 188 m from where the front of the train started. What will be the speed of the last car as it passes the worker?
vi = 0
vf = 18
d = 188
a = ? T
The equation is vf^2 = vi^2 + 2ad
Solve for a = (vf^2 - vi^2) divided by 2d
= (18^2 - 0^2) / 2 x 188 = .86 m/s^2
If the acceleration continues for the additional 90 m.
The new list is: vi = 18
vf = ?
d = 90
a = .86
vf^2 = vi^2 + 2ad Solving for vf = square root of vi^2 + 2ad
= square root of (18^2 + 2 x .86 x 90)
= 21.9 m/s