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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?

Sagot :

jacob193

Answer:

[tex]9.5\; {\rm m}[/tex].

Explanation:

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

The duration of this acceleration is [tex]t = 2\; {\rm s}[/tex].  

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

[tex]\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}[/tex].

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

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