Question

1. 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.
2. a powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s.

Explanation:

Step1: First find the acceleration

We use the formula $a=\frac{v - u}{t}$, where $v = 40.0\ m/s$ (final - velocity), $u = 8.00\ m/s$ (initial - velocity), and $t=3.33\times 10^{-2}\ s$ (time).
$a=\frac{40.0 - 8.00}{3.33\times 10^{-2}}=\frac{32.0}{3.33\times 10^{-2}}\ m/s^{2}\approx961\ m/s^{2}$

Step2: Then find the distance

We use the formula $s=ut+\frac{1}{2}at^{2}$.
Substitute $u = 8.00\ m/s$, $a\approx961\ m/s^{2}$, and $t = 3.33\times 10^{-2}\ s$ into the formula.
$s=(8.00\times3.33\times 10^{-2})+\frac{1}{2}\times961\times(3.33\times 10^{-2})^{2}$
First term: $8.00\times3.33\times 10^{-2}=0.2664\ m$
Second - term: $\frac{1}{2}\times961\times(3.33\times 10^{-2})^{2}=\frac{1}{2}\times961\times1.10889\times 10^{-3}\approx0.530\ m$
$s=0.2664 + 0.530=0.7964\ m\approx0.80\ m$

Answer:

$0.80\ m$