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. 236.)

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