Question

A retarding force is applied to stop a motorcar. If the speed of the motorcar is doubled, how much more distance will it cover before stopping under the same retarding force?

Answer 1

Step 1: Given

A retarding force is applied to stop a motorcar. If the speed of the motorcar is doubled

Step 2:

From the work-energy theorem, we have
Work done against retarding force = Loss in K.E.
In the first case, we have
$F\times x=\frac{1}{2}m{v}^2\dots \left(1\right)$
In the second case, we have
$F\times x^{\text{'}}=\frac{1}{2}m{\left(v^{\text{'}}\right)}^2$
$=\frac{1}{2}m(2v{)}^2\left(v^{\text{'}}=2v\right)$
$F\times x^{\text{'}}=\frac{1}{2}m\times 4v^2\dots \left(2\right)$
$F\times x^{\text{'}}=4\times \frac{1}{2}m{v}^2$
Using (1), we have
$F{x}^{\text{'}}=4Fx$
$F{x}^{\text{'}}=F4x$
$\therefore x^{\text{'}}=4x$
Therefore the motor car will cover a distance four times longer than before.