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Question: Answered & Verified by Expert
A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time ' $t$ ' is proportional to
PhysicsWork Power EnergyJEE MainJEE Main 2003
Options:
  • A
    $t^{3 / 4}$
  • B
    $t^{3 / 2}$
  • C
    $t^{1 / 4}$
  • D
    $t^{1 / 2}$
Solution:
1832 Upvotes Verified Answer
The correct answer is:
$t^{3 / 2}$
Power $=$ F. V
$\mathrm{F}=\mathrm{m}\left(\frac{\mathrm{dV}}{\mathrm{dt}}\right) \Rightarrow \mathrm{m} \cdot \mathrm{v} \cdot \frac{\mathrm{dV}}{\mathrm{dt}}=$ constant $=\mathrm{C}$
$\Rightarrow \frac{\mathrm{dV}}{\mathrm{dt}}=\frac{\mathrm{C}}{\mathrm{m}}=\mathrm{k} \Rightarrow \mathrm{vdv}=\mathrm{kdt} \quad \Rightarrow \int \mathrm{vdv}=\int \mathrm{kdt} \Rightarrow \frac{\mathrm{V}^2}{2}=\mathrm{kt}+\mathrm{c}$
$\Rightarrow \mathrm{v} \propto(\mathrm{t})^{1 / 2}$ $\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{c} \cdot \mathrm{t}^{1 / 2}$
$\Rightarrow \int \mathrm{ds}=\int\left(\mathrm{c} \cdot \mathrm{t}^{1 / 2}\right) \mathrm{dt} \Rightarrow \mathrm{S}=\mathrm{C} \cdot \frac{2}{3} \mathrm{t}^{3 / 2}$ $\Rightarrow \mathrm{S}=\frac{\mathrm{ct}^{3 / 2}}{3 / 2} \Rightarrow \mathrm{s} \propto \mathrm{t}^{3 / 2}$

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