Question

A 95-m-long train begins uniform acceleration from rest. the front of the train has a speed of 25 m/s when it passes a railway worker who is standing 180 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. 2–38)

Answer 1

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Answer 2

Hey dear,

◆ Answer-
v = 30.9 m/s

◆ Explaination-
# Given-
train length = 95 m
u = 0 m/s

# Solution-
Let a be the uniform acceleration.

When front of train bypass the observer,
v = 25 m/s at s = 180 m

Newton's 3rd kinematic eqn,
v^2 = u^2 + 2as
25^2 = 0 + 2×a×180
a = 625/360
a = 1.736 m/s^2

When end train bypass the observer,
a = 1.736 m/s^2, s = 180+95 = 275 m

Newton's 3rd kinematic eqn,
v^2 = u^2 + 2as
v^2 = 0 + 2 × 1.736 × 275
v^2 = 955
v = 30.9 m/s

Therefore, speed of the train when its end bypass the observer is 30.9 m/s.

Hope this helps you...