Stopping of Vehicle by Retarding Force
Category : JEE Main & Advanced
If a vehicle moves with some initial velocity and due to some retarding force it stops after covering some distance after some time.
(1) Stopping distance : Let m = Mass of vehicle,
\[\upsilon \] = Velocity, P = Momentum, E = Kinetic energy
F = Stopping force, x = Stopping distance,
t = Stopping time
Then, in this process stopping force does work on the vehicle and destroy the motion.
By the work- energy theorem
\[W=\Delta K=\frac{1}{2}m{{v}^{2}}\]
\[\Rightarrow \] Stopping force (F) x Distance (x) = Kinetic energy (E)
\[\Rightarrow \] Stopping distance (x) \[=\frac{\text{Kinetic}\,\text{energy}\,\text{(}E\text{)}}{\text{Stopping}\,\text{force}\,\text{(}F\text{)}}\]
\[\Rightarrow \] \[x=\frac{m{{v}^{2}}}{2F}\] ...(i)
(2) Stopping time : By the impulse-momentum theorem
\[F\times \Delta t=\Delta P\Rightarrow F\times t=P\]
\[\therefore \] \[t=\frac{P}{F}\]
or \[t=\frac{mv}{F}\] ...(ii)
(3) Comparison of stopping distance and time for two vehicles : Two vehicles of masses \[{{m}_{1}}\] and \[{{m}_{2}}\] are moving with velocities \[{{\upsilon }_{1}}\] and \[{{\upsilon }_{2}}\] respectively. When they are stopped by the same retarding force (F).
The ratio of their stopping distances \[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{m}_{1}}v_{1}^{2}}{{{m}_{2}}v_{2}^{2}}\] and the ratio of their stopping time \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{m}_{1}}{{v}_{1}}}{{{m}_{2}}{{v}_{2}}}\]
(i) If vehicles possess same velocities
\[{{\upsilon }_{1}}={{\upsilon }_{2}}\]
\[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\] ; \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\]
(ii) If vehicle possess same kinetic momentum
\[{{P}_{1}}={{P}_{2}}\]
\[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}=\left( \frac{P_{1}^{2}}{2{{m}_{1}}} \right)\,\left( \frac{2{{m}_{2}}}{P_{2}^{2}} \right)\,=\frac{{{m}_{2}}}{{{m}_{1}}}\]
\[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{P}_{1}}}{{{P}_{2}}}=1\]
(iii) If vehicle possess same kinetic energy
\[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}=1\]
\[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{\sqrt{2{{m}_{1}}{{E}_{1}}}}{\sqrt{2{{m}_{2}}{{E}_{2}}}}=\sqrt{\frac{{{m}_{1}}}{{{m}_{2}}}}\]
Note :
If vehicle is stopped by friction then
Stopping distance \[x=\frac{\frac{1}{2}m{{v}^{2}}}{F}\]\[=\frac{\frac{1}{2}m{{v}^{2}}}{ma}\]\[=\frac{{{v}^{2}}}{2\mu g}\]
\[[\text{As}\,\,a=\mu g]\]
Stopping time \[t=\frac{mv}{F}\]\[=\frac{mv}{m\mu \,g}\]\[=\frac{v}{\mu \,g}\]
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