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

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

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

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Answer 1 · 2016-04-15T18:14:48

Minimum stopping distance = 45.49m

Explanation:

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

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

so dividing both side by $m$ $m$ gives:

$μ g = a = 0.7 \cdot 9.81 = 6.87 m s^{- 2}$ $mug=a=0.7*9.81=6.87ms^-2$

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

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

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

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

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

1 Answer

Explanation:

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1 Answer

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