What would be the stopping distance if the ve | vedclass

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(C) The stopping distance $s$ of a vehicle is given by the work-energy theorem,where the work done by the braking force $F$ equals the change in kinet

Vedclass · Accepted Answer

$9$

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(C) The stopping distance $s$ of a vehicle is given by the work-energy theorem,where the work done by the braking force $F$ equals the change in kinetic energy: $F \cdot s = \frac{1}{2}mv^2$. Since the braking force $F$ and mass $m$ are constant,the stopping distance is proportional to the square of the velocity: $s \propto v^2$. If the initial velocity is $v_1 = v$ and the stopping distance is $s_1 = d$,then for a new velocity $v_2 = 3v$,the new stopping distance $s_2$ is: $\frac{s_2}{s_1} = \left(\frac{v_2}{v_1}\right)^2 = \left(\frac{3v}{v}\right)^2 = 3^2 = 9$. Therefore,$s_2 = 9d$.

What would be the stopping distance if the velocity of a vehicle becomes three times (in $d$)?

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