Question

A retarding force is applied to stop the motor if the speed of motor car is doubled how much more will it cover before stopping under the same retarding force

Answer 1

Case -I
initial velocity = u
final velocity, v = 0

Let the retarding force = F
mass of car = m
retardation = -a = F/m

Distance travelled before stopping,
$S_1 = \frac{ {v}^{2} -{u}^{2} }{2 \times ( - a) } \\ \\ S_1 = \frac{0 - {u}^{2} }{ - 2a} \\ \\ S_1= \frac{ {u}^{2} }{2a}$
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Case - II.

initial velocity is twice that of case I

initial velocity = 2u
final velocity, v = 0

retarding force (remains same ) = F
mass of car = m
retardation = -a = F/m

Distance travelled before stopping,

$S_2= \frac{ {v}^{2} -{(2u)}^{2} }{2 \times ( - a) } \\ \\ S_2 = \frac{0 - {(2u)}^{2} }{ - 2a} \\ \\ S_2= \frac{ 4{u}^{2} }{2a} \\ \\ \boxed{S_2=4 S_1}$

Thus extra distance travelled
= S2 - S1

= 4 S1 - S1

= 3 S1

where S1 is the distance travelled in the first case.