Question

A train moves from rest to a speed of 45m/s in 65s. What is its acceleration

Answer 1

0.69 m/s²

Step-by-step explanation:

Acceleration is change in velocity over change in time which can be expressed as:

`\displaystyle{a = (\Delta v)/(\Delta t) = (v_2-v_1)/(t_2-t_1)}`

In this case,
`v_2` is final velocity which is 45 m/s and
`v_1` is 0 m/s because it's said that the train moves from rest meaning the initial velocity is 0 m/s.
`t_2` is final time which is 65 seconds and the initial time is 0 second. Therefore:

`\displaystyle{a = (45-0)/(65-0)}\\\\\displaystyle{a=(45)/(65)}\\\\\displaystyle{a=0.69 \ \: \text{m/s}^2}`

Therefore, the acceleration is 0.69 m/s²