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(B) Let the initial mass of the vehicle be $m_1 = m$. The initial velocity is $u$ and the final velocity is $v = 0$. Using the equation of motion $v^2

Vedclass · Accepted Answer

$1.4d$

Vedclass · Answer

(B) Let the initial mass of the vehicle be $m_1 = m$. The initial velocity is $u$ and the final velocity is $v = 0$. Using the equation of motion $v^2 = u^2 + 2aS$,we have $0 = u^2 - 2ad$,which gives the retardation $a = \frac{u^2}{2d}$.Since the retardation $a$ is constant,the retarding force $F = m_1 a = m \left(\frac{u^2}{2d}\right)$.In the second case,the mass becomes $m_2 = m + 0.4m = 1.4m$. The retardation $a$ remains the same.The new stopping distance $d'$ is given by $d' = \frac{u^2}{2a}$.Substituting $a = \frac{u^2}{2d}$,we get $d' = \frac{u^2}{2(u^2/2d)} = d$.However,if the retarding force $F$ is constant (as implied by the context of braking force),then $F = m_1 a_1 = m_2 a_2$. Since $F$ is constant,$a_2 = \frac{F}{m_2} = \frac{ma}{1.4m} = \frac{a}{1.4}$.Then $d' = \frac{u^2}{2a_2} = \frac{u^2}{2(a/1.4)} = 1.4 \left(\frac{u^2}{2a}\right) = 1.4d$.

$A$ vehicle without passengers is moving on a frictionless horizontal road with velocity $u$ and can be stopped in a distance $d$. Now,$40\%$ of its weight is added. If the retardation remains the same,what is the stopping distance at velocity $u$?

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