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27 PE

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Found in: Page 83

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes$${\bf{3}}{\bf{.33 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}\;{\bf{s}}$$, calculate the distance over which the puck accelerates.

The distance for acceleration of puck is $$0.8\;{\rm{m}}$$.

See the step by step solution

Determination of acceleration of puck

Given Data:

The initial velocity of racer is $$u = 8\;{\rm{m}}/{\rm{s}}$$

The final velocity of racer is $$v = 40\;{\rm{m}}/{\rm{s}}$$

The time for acceleration of slap shot is $$t = 3.33 \times {10^{ - 2}}\;{\rm{s}}$$

The distance travelled by puck is found by first calculating the acceleration by first equation of motion and then applying the third equation of motion.

The acceleration of the puck is given as

$$v = u + at$$

Here, $$a$$ is the deceleration of the puck.

Substitute all the values in the above equation.

$$\begin{array}{c}40\;{\rm{m}}/{\rm{s}} = 8\;{\rm{m}}/{\rm{s}} + a\left( {3.33 \times {{10}^{ - 2}}\;{\rm{s}}} \right)\\a = 961\;{\rm{m}}/{{\rm{s}}^2}\end{array}$$

Determination of distance for acceleration of puck

The distance for acceleration of the puck is given as

$${v^2} = {u^2} + 2ad$$

Substitute all the values in the above equation.

$\begin{array}{c}{\left( {40\;{\rm{m}}/{\rm{s}}} \right)^2} = {\left( {8\;{\rm{m}}/{\rm{s}}} \right)^2} + 2\left( {961\;{\rm{m}}/{{\rm{s}}^2}} \right)d\\d = 0.8\;{\rm{m}}\end{array}$

Therefore, the distance for acceleration of the puck is $$0.8\;{\rm{m}}$$.