Question

A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time 't' is proportional to :

• A. t³/2
• B. t¹/2
• C. t²/3
• D. t

Hint:

For constant power $P$, remember these shortcuts: $v \propto t^{1/2}$, $s \propto t^{3/2}$, and $a \propto t^{-1/2}$.

Answer 1

The Correct Option is A

Step 1: Power $P = Fv = (ma)v = m \frac{dv}{dt} v$.
Step 2: $P dt = m v dv \implies Pt = \frac{1}{2} m v^2 \implies v = \sqrt{\frac{2P}{m}} t^{1/2}$.
Step 3: $v = \frac{ds}{dt} = k t^{1/2} \implies s = \int k t^{1/2} dt = k' t^{3/2}$. Therefore, $s \propto t^{3/2}$.