Question

The hockey player is moving at a speed of 9. 5 m/s. if it takes him 2 seconds to come to a stop under constant acceleration, how far does he travel while stopping?

Answer 1

`9.5\; {\rm m}`.

Step-by-step explanation:

Let
`u` and
`v` denote the velocity of this hockey player before and after stopping, respectively. The question states that
`u = 9.5\; {\rm m\cdot s^(-1)}` and implies that
`v = 0\; {\rm m\cdot s^(-1)` since the hockey player has come to a stop.

The duration of this acceleration is
`t = 2\; {\rm s}`.

Since the acceleration of this hockey player was constant, SUVAT equation would apply. In particular, the SUVAT equation
`x = (1/2)\, (v + u) \, (t)` gives the displacement
`x` of this hockey player during that
`2\; {\rm s}` of acceleration:

`\begin{aligned} x &= (1)/(2)\, (9.5\; {\rm m\cdot s^(-1)} + 0\; {\rm m\cdot s^(-1)})\, (2\; {\rm s}) = 9.5\; {\rm m} \end{aligned}`.

In other words, this hockey player would have travelled
`9.5\; {\rm m}` while stopping.