WBJEE · Physics · Work Power Energy
A box is moved along a straight line by a machine delivering constant power. The distance moved by the body in time \(\mathrm{t}\) is proportional to
- A \(\mathrm{t}^{\frac{1}{2}}\)
- B \(t^{\frac{3}{4}}\)
- C \(\mathrm{t}^{\frac{3}{2}}\)
- D \(\mathrm{t}^2\)
Answer & Solution
Correct Answer
(C) \(\mathrm{t}^{\frac{3}{2}}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text {Hints: } \mathrm{P}=\mathrm{Fv}=\mathrm{m} \cdot \frac{\mathrm{dv}}{\mathrm{dt}} \cdot \mathrm{v} \\ & \int \mathrm{vdv}=\int \mathrm{P} / \mathrm{mdt} ; \quad \frac{\mathrm{v}^2}{2}=\frac{\mathrm{Pt}}{\mathrm{m}} \\ & V=\sqrt{\frac{2 p}{m}}…
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