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.

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