Question

Train is traveling at an initial velocity of 68.325m/s. After 23.75 seconds it speeds up to a final velocity of 79.32m/s. What is the train's acceleration during this time

Answer 1

So,

We know that:

`acceleration = ( final\ velocity\ -\ initial\ velocity)/(time)`

Plug in the values.

`acceleration = (79.32 m/s^(2)\ -\ 68.325m/s^(2))/(23.75 secs)`

`acceleration = (10.995m/s^(2) )/(23.75 secs)`

`acceleration = 0.4629 m/s^(2)`