First and foremost is knowing what information you need and what you don't. The question asks about the acceleration during the first 5.6 km of travel, so that is all we really care about for this problem.
The problem says the train starts from rest (0 m/s) and accelerates uniformly (constantly) until 5.6km and has acquired a velocity of 42 m/s. Acceleration is change in velocity over change in time. We know the initial speed, the final speed and the total distance traveled. However, to find acceleration we need the time.
To find the time we need to know the general equation of motion for constant acceleration. 0.5*a*t^2+v*t+s=position. a is the acceleration, t is the time, v is the initial velocity and s is the initial starting position. If we 'plug in' what we know (initial velocity is zero and initial position is zero), we have 0.5*a*t^2=position. We also know that acceleration changes from 0 to 42 m/s. We still don't know the time, but let us write it as an equation. This means the acceleration becomes (42-0)/t = 42/t = a. If we put this back into our equation we find position = 0.5*42/t*t^2 = 21*t. Knowing the position of 5.6 km, or 5,600 meters, we can find t.
5600=21*t results in t=266.67 s. We then go back to our equation for acceleration and 'plug in' t. This results in a= 42/t or 42/266.67 = 0.1575 m/s^2 (v= m/s and t=s, m/s/s =m/s^2). If we round, we get 0.16 m/s^2.
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