Question

3. A train is 120 m long. Sydney was standing 200 m from the front of the train when it

began to accelerate from rest. She noticed that the front of the train was moving at 8.00
m/s when it passed her. How fast would the back of the train be going when it passed
her?

Answer 1

Below

Step-by-step explanation:

Find acceleration first

vf = vo + at (starts from rest so vo = 0)

8 = 0 + at

8 = at

d = vo t + 1/2 a t^2 (substitute 8 in for 'at')

200 = 0 t + 1/2 (8) t

200 / 4 = t = 50 seconds for front of train to pass her at 8 m/s

acceleration = change in velocity / change in time = 8 / 50 = .16 m/s^2

Back of train has to travel 120 meters at initial v = 8 m/s and accelerating at .16 m/s^2 :

d = vo t + 1/2 a t^2

120 = 8 t + 1/2 ( .16 ) t^2

.16t^2 + 8t - 120 = 0 Quadratic Formula shows t = 13.25 s

vf = vo + a t

vf = 8 m/s + .16 m/s^2 ( 13.25 s) = 10.1 m/s