A vehicle of mass m is moving on a rough horizontal road with momentum p. If the coefficient of friction between the tyres and the road be u, then the stopping distance is given by..??

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A vehicle of mass m is moving on a rough horizontal road with momentum p. If the coefficient of friction between the tyres and the road be u, then the stopping distance is given by..??

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Answer 1 · 2018-04-11T10:47:57

Initial velocity $u = \frac{p}{m}$ $u=p/m$
Final velocity $v = 0$ $v=0$ , (as the vehicle must stop)
Force of friction $= μ m g$ $=mumg$
(where $g$ $g$ is acceleration due to gravity.)
Acceleration due to friction $= - \frac{μ m g}{m} = - μ g$ $= -(mumg)/m=-mug$
( $-$ $-$ ve sign shows that it is retardation.)

Using the kinematic expression

$v^{2} - u^{2} = 2 a s$ $v^2-u^2=2as$

and inserting various values we get stopping distance $s$ $s$

${(0)}^{2} - \frac{p^{2}}{m^{2}} = 2 (- μ g) s$ $(0)^2-p^2/m^2=2(-mug)s$
$\Rightarrow s = \frac{p^{2}}{2 m^{2} μ g}$ $=>s=p^2/(2m^2mug)$

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A vehicle of mass m is moving on a rough horizontal road with momentum p. If the coefficient of friction between the tyres and the road be u, then the stopping distance is given by..??

1 Answer

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