MixedJEE Physics · Original learning card5 original chapter questions

Constant-Power Motion

For one-dimensional constant-mass motion under constant net power, P=mv(dv/dt), giving v^2=v_0^2+2Pt/m and displacement from integrating that speed.

Why this shows up in the exam

Constant-power engines · Time scaling of speed · Distance-time power laws

Learn the idea

Constant power produces acceleration that falls as speed rises. When a machine supplies fixed power, the same energy arrives each second, so speed grows like square root of time rather than linearly.

🧠 Memory hook: Fixed power adds equal kinetic energy each second.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • P = mv dv/dt — constant-mass one-dimensional power
  • v² = v₀²+2Pt/m — speed under constant net power
  • x-x₀ = integral v dt — displacement after finding speed

How to approach it

  1. 1Write dK/dt=P
  2. 2Integrate for v(t)
  3. 3Integrate again only if displacement is asked

Common slip-ups that cost marks

  • •Assuming constant power means constant force
  • •Using v proportional to t
  • •Forgetting nonzero initial speed

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 5

A 2 kg body speeds up from 3 m/s to 7 m/s. What net work is done on it?

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