Question

Until a train is a safe distance from the station, it must travel at 5 m/s. Once the train is on open track, it can speed up

Answer 1

Answer:

The acceleration of the train is 5 m/s².

Explanation:

Acceleration is the change in velocity over a given time interval.

Therefore, acceleration is given as:

$a=\frac{v_{f}-v_{i}}{\Delta t}$

Where, $v_{i}\textrm{ and }v_{f}$ are initial and final velocities and $\Delta t$ is the time interval.

Here, $v_{i}=5\textrm{ m/s},v_{f}=45\textrm{ m/s},\Delta t=8\textrm{ s}$

This gives,

$a=\frac{45-5}{8}=\frac{40}{8}=5\textrm{ }m/s^{2}$

Therefore, the acceleration of the train is 5 m/s².