How do you determine the minimum stopping distance of a motorcycle, given its velocity and the kinetic coefficient of friction?

A motorcycle moving at $25.0 \frac{m}{s}$ slides to a stop. Calculate the minimum stopping distance if the kinetic coefficient of friction between the tire and the road is $0.7$.

Apr 15, 2016

Minimum stopping distance = 45.49m

Explanation:

The friction between the bike and the road will be $\mu R$ where $\mu$ is the coefficient of friction and $R$ is the normal reaction between the bike and the road. $R$ will equal the weight of the bike, $m g$, so friction =$\mu m g$

If this is the only force acting on the bike, then by Newton's 2nd law
$F = m a$
$\mu m g = m a$

so dividing both side by $m$ gives:

$\mu g = a = 0.7 \cdot 9.81 = 6.87 m {s}^{-} 2$

We can now use the equation of motion:
${v}^{2} = {u}^{2} + 2 a s$

We know the initial velocity, $u = 25 m {s}^{-} 1$, acceleration $a = - 6.87 m {s}^{-} 2$ (negative since it is decelerating), and $v$, final velocity is $0 m {s}^{-} 1$ (since the bike comes to rest). We want to find the distance, $s$.

So
$0 = {25}^{2} - 2 \cdot 6.87 . s$
$0 = 625 - 13.74 \cdot s$
$s = \frac{625}{13.74} = 45.49 m$