A bus moving on a level road with a velocity | vedclass

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Question

(A) Using the work-energy theorem or the equation of motion $v^2 - u^2 = 2as$,where $v = 0$ (final velocity),$u = v$ (initial velocity),and $a = -F/m$

Vedclass · Accepted Answer

$1.25 x$

Vedclass · Answer

(A) Using the work-energy theorem or the equation of motion $v^2 - u^2 = 2as$,where $v = 0$ (final velocity),$u = v$ (initial velocity),and $a = -F/m$ (retardation).$0^2 - v^2 = 2(-F/m)x$$v^2 = 2Fx/m$$x = (mv^2) / (2F)$Since $v$ and $F$ are constant,the stopping distance $x$ is directly proportional to the mass $m$ of the bus $(x \propto m)$.Let the initial mass be $m_1 = m$ and the initial distance be $x_1 = x$.The new mass after a $25 \%$ increase is $m_2 = m + 0.25m = 1.25m$.Using the proportionality $x_2 / x_1 = m_2 / m_1$:$x_2 / x = (1.25m) / m$$x_2 = 1.25x$.

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