A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a velocity of 42m/s What is the approximate acceleration of the train during this time?

The solution to the problem is as follows:

42 m/s * 1000m/km =.042 km/s

a = (vf - vi) / t D = vit+1/2at^2 (vit can be cancelled)

5.6km = 1/2 ((vf-vi) / t) * t^2

5.6km = 1/2 ((.042km/s-0km/s) / t) * t^2

5.6km = 1/2 (.042km/s/t) * t^2

5.6km = (.021km/s) / t * t^2

5.6km =.021km/s * t

t = 266.666667s

checking work:

a = (vf-vi) / t

a = (.042km/h-0km/h) / 266.6667s

a =.001575km/s^2

D = vit + 1/2at^2 56km = 1/2 (.001575 km/s^2) * (266.6667s) ^2

56km =.0007875 km/s^2 * 71111.12889s^2

56km = 56km