Question

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 3.33×10 −2 s, calculate the distance over which the puck accelerates.

Answer 1

We use the equations of motion,

$v = u + at$                                               (A)

$v^{2} = u^{2} + 2as$                                (B)

Here, v is final velocity of the body, u is initial velocity of the body, a is the acceleration of the body, t is the time taken in the motion and s is distance traveled by the body.

Given $u = 8 \ m/s$, $v = 40 \ m/s$ and $t = 3.33 \times 10^{-2} \ s$.

So, from equation (A)

$40 \ m/s = 8 \ m/s + a ( 3.33 \times 10^{-2 } \ s) \\\\ a =  \frac{32 \ m/s}{3.33\times 10^{-2 } \ s  }  = 9.6 \times 10^{2} \ m/s^2$

Now from the  equation (B),

$(40 \ m/s)^2 = (8 \ m/s)^2 + 2 (9.6 \times 10^{2} m/s^2) \times s  \\\\ s = \frac{(40 \ m/s)^2-(8 \ m/s)^2}{2 (9.6 \times 10^{2} m/s^2} = 0.8 \ m$

Thus, the distance over which the puck accelerates is 0.8 m.