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 :-

Solution

Physics sir · 3 steps

Constant power means the box's kinetic energy grows linearly with time, so start from there.

1. Work equals energy. In time t the machine does work Pt. The box starts from rest on a horizontal line, so all of this work becomes kinetic energy.

   Pt = 1/2mv² ⇒ v = √(2P/m) t^(1/2)

2. Speed to distance. So v ∝ t^(1/2). Distance is the area under the v–t graph, so integrate: s = ∫₀ᵗ √(2P/m) t^(1/2) dt = 2/3√(2P/m) t^(3/2).
3. So. The distance moved is proportional to t^(3/2), which is option (b).

The idea to keep. For constant power from rest, use Pt = 1/2mv². It gives v ∝ t^(1/2), and integrating once more gives s ∝ t^(3/2).

Answer from the NTA final answer key. Solution by PrepWiser.