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

Answer:

$9.5\; {\rm m}$.

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:

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

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