Posted by **Sydney** on Tuesday, September 13, 2011 at 10:05pm.

A 90 m long train begins uniform acceleration from rest. The front of the train has a speed of 21 m/s when it passes a railway worker who is standing 184 m from where the front of the train started. What will be the speed of the last car as it passes the worker? (See Fig. 2-36.)

- Physics -
**drwls**, Tuesday, September 13, 2011 at 10:22pm
First calculate the acceleration of the front of the train. When its velocity is V after traveling a distance X from a standing start,

V^2 = 2 a X, so

a = V^2 /(2X) = 1.198 m/s^2

When the last car passes the worker, the first car will have traveled 184 + 90 = 274 m

Use again V^2 = 2 a X, with the new value of X = 274. (a remains the same). This new V is the velocity of both the first and the last car at that time.

V^2 = 2*1.198*274

V = 25.6 m/s

- Physics -
**Anonymous**, Wednesday, September 19, 2012 at 4:46pm
seriously bruh

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