Question

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?

Answer 1

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