Question

Initially stationary, a train has a constant acceleration of 1.8 m/s²  

How far it  travel after 14.2s.  Calculate answer to one decimal.

Answer 1

As per question, initial velocity of train, u = 0 m/s, acceleration of the train, a = 1.8 m/s^2

To find: s = distance (m)=?
t = time (s) = 14.2 s

For constant acceleration, the equation of motion we know, v = u + at;............................(1)

s = ut + (1/2)(a)(t^2);.....................................(2)

v^2 = u^2 + 2as............................................(3)

so we can use two methods:

1. we can find v, then put it in (3) to get the distance the train travels or

2. Directly find it by using (2).

Now by using (1) v = u + at , u=0 (since initially it was at stationary position),a=1.8m/s^2, t=14.2s

v = 0 +(1.8 m/s^2)(14.2 s) = 25.56 m/s

Then by (3), s = (25.56 m/s)^2 / (2*1.8 m/s^2) = 181.5 m

Alternatively, using (2),

s = (0 m)(14.2 s) + (1/2*(1.8 m/s^2) (14.2 s)^2) =181.5 m

Hence, After 14.2 s train has travelled 181.5 meters distance.