Position and Velocity of an Automobile w.r.t Time
Category : JEE Main & Advanced
An automobile of mass m accelerates, starting from rest, while the engine supplies constant power P, its position and velocity changes w.r.t time.
(1) Velocity : As \[F\upsilon =P=\] constant
i.e. \[m\frac{dv}{dt}v=P\] \[\left[ \text{As }F=\frac{mdv}{dt} \right]\]
or \[\int_{{}}^{{}}{v\,dv}=\int_{{}}^{{}}{\frac{P}{m}dt}\]
By integrating both sides we get \[\frac{{{v}^{2}}}{2}=\frac{P}{m}t+{{C}_{1}}\]
As initially the body is at rest i.e. \[\upsilon =0\] at t = 0, so \[{{C}_{1}}=0\]
\[\therefore \] \[v={{\left( \frac{2Pt}{m} \right)}^{1/2}}\]
(2) Position : From the above expression \[v={{\left( \frac{2Pt}{m} \right)}^{1/2}}\]
or \[\frac{ds}{dt}={{\left( \frac{2Pt}{m} \right)}^{1/2}}\]
\[\left[ \text{As }v=\frac{ds}{dt} \right]\]
i.e. \[\int_{{}}^{{}}{ds}=\int_{{}}^{{}}{{{\left( \frac{2Pt}{m} \right)}^{1/2}}\,dt}\]
By integrating both sides we get
\[s={{\left( \frac{2P}{m} \right)}^{1/2}}.\frac{2}{3}{{t}^{3/2}}+{{C}_{2}}\]
Now as at t = 0, s = 0, so \[{{C}_{2}}=0\]
\[s={{\left( \frac{8P}{9m} \right)}^{1/2}}{{t}^{3/2}}\]
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