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27 PE

Expert-verifiedFound in: Page 83

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 \(0.8\;{\rm{m}}\).

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.

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

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 \(0.8\;{\rm{m}}\).

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