Work Energy Power

JEE Main level, single correct, medium

167 of 183

Question 167 of 183

A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time ' tt ' is proportional to

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Answer: (B) t3/2t^{3 / 2}

Power == F. V
F=m(dVdt)⇒m⋅v⋅dVdt=\mathrm{F}=\mathrm{m}\left(\frac{\mathrm{dV}}{\mathrm{dt}}\right) \Rightarrow \mathrm{m} \cdot \mathrm{v} \cdot \frac{\mathrm{dV}}{\mathrm{dt}}= constant =C=\mathrm{C}
⇒dVdt=Cm=k⇒vdv=kdt⇒∫vdv=∫kdt⇒V22=kt+c\Rightarrow \frac{\mathrm{dV}}{\mathrm{dt}}=\frac{\mathrm{C}}{\mathrm{m}}=\mathrm{k} \Rightarrow \mathrm{vdv}=\mathrm{kdt} \quad \Rightarrow \int \mathrm{vdv}=\int \mathrm{kdt} \Rightarrow \frac{\mathrm{V}^2}{2}=\mathrm{kt}+\mathrm{c}
⇒v∝(t)1/2\Rightarrow \mathrm{v} \propto(\mathrm{t})^{1 / 2} dsdt=c⋅t1/2\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{c} \cdot \mathrm{t}^{1 / 2}
⇒∫ds=∫(c⋅t1/2)dt⇒S=C⋅23t3/2\Rightarrow \int \mathrm{ds}=\int\left(\mathrm{c} \cdot \mathrm{t}^{1 / 2}\right) \mathrm{dt} \Rightarrow \mathrm{S}=\mathrm{C} \cdot \frac{2}{3} \mathrm{t}^{3 / 2} ⇒S=ct3/23/2⇒s∝t3/2\Rightarrow \mathrm{S}=\frac{\mathrm{ct}^{3 / 2}}{3 / 2} \Rightarrow \mathrm{s} \propto \mathrm{t}^{3 / 2}

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