JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Position and Velocity of an Automobile w.r.t Time

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 \[{{C}_{2}}=0\]           

\[s={{\left( \frac{8P}{9m} \right)}^{1/2}}{{t}^{3/2}}\]