Physics
posted by Sydney .
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.)

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 
seriously bruh
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