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Most kinematics you learn assumes constant acceleration. An engine at full throttle or a motor hauling a load behaves differently: it delivers roughly constant power, so the force it supplies changes with speed. Students who reach for v = u + at here get wrong answers that look reasonable. This post shows how to set up constant-power motion, works two examples with numbers you can check, and lists the traps.
Why do the usual equations of motion fail at constant power?
The NCERT Class 11 chapter Work, Energy and Power defines instantaneous power as P = dW/dt and shows that it equals F·v, the force times the velocity of the body it acts on. If P is fixed, the driving force is F = P/v. As speed rises, force falls, and so does the acceleration a = P/(mv). The equations v = u + at and v² = u² + 2as assume a constant acceleration, so they do not apply.
The Syllabus for JEE (Main) 2026 lists power and "work done by a constant force and a variable force" in Unit 4, and the JEE (Advanced) 2026 brochure lists work and power under mechanics. Check the bulletin for your exam year.
How do you get speed and distance from constant power?
Use energy, not force. With no resistance, all the engine's work goes into kinetic energy. Starting from rest, after time t:
½mv² = Pt, so v = √(2Pt/m).
For distance, write the force as m v dv/dx: P/v = m v dv/dx, so P dx = m v² dv. Integrating from rest gives x = mv³/(3P).
The second form is the one to remember, because many questions give a speed or a distance and never mention time. If you want the integration steps in more detail, our guide to the calculus you need for JEE physics covers the a = v dv/dx trick.
Worked example 1. A 1000 kg car starts from rest on a level road. Its engine delivers a constant 20 kW to the wheels. Ignore friction and air resistance. Find its speed, acceleration and distance travelled after 10 s.
Speed: v² = 2Pt/m = 2 × 20,000 × 10 / 1000 = 400, so v = 20 m/s.
Acceleration at that moment: a = P/(mv) = 20,000 / (1000 × 20) = 1 m/s².
Distance: x = mv³/(3P) = 1000 × 8000 / 60,000 ≈ 133 m.
Check by a second route: integrating v = √(2P/m) × √t gives x = (2/3)√(2P/m) t^(3/2) = (2/3) × √40 × 10^1.5 ≈ (2/3) × 6.32 × 31.6 ≈ 133 m. Both agree.
Compare with constant acceleration: a car reaching 20 m/s in 10 s at a steady rate would cover only 100 m. At constant power the acceleration is largest at the start, so the car covers more ground early.
One caution: the formula predicts a very large acceleration as v approaches zero. A real car cannot do that, because tyre friction limits the force. "Constant power from rest" is an idealisation that examiners use on purpose.
How do proportionalities speed up MCQs?
From the two formulas:
v ∝ t^(1/2)
x ∝ t^(3/2)
v ∝ x^(1/3)
a ∝ 1/v
Worked example 2. For the same car, how long does it take to reach 30 m/s, and how far does it go? Time: t = mv²/(2P) = 1000 × 900 / 40,000 = 22.5 s. Distance: x = 1000 × 27,000 / 60,000 = 450 m. Check with ratios: speed went up 1.5 times, so time should rise 1.5² = 2.25 times (10 s to 22.5 s) and distance 1.5³ ≈ 3.375 times (133 m to 450 m). Both match.
What changes when there is resistance?
With a resistive force f, the net force is P/v − f. Top speed is reached when the two balance: v_max = P/f. For 20 kW against 500 N, v_max = 40 m/s. At 20 m/s the driving force is 20,000/20 = 1000 N, so the acceleration is (1000 − 500)/1000 = 0.5 m/s². If the resistance is proportional to speed, kv, then the balance P/v = kv gives v_max = √(P/k). Our post on power and efficiency works through lifts, pumps and top speed in more detail.
Why do students find this hard? In a study by Lawson and McDermott (1987), students compared the changes in momentum and kinetic energy of two pucks pushed by the same constant force; most could not relate the algebra they had learned to the motion they watched. Constant power adds a twist, because the force is not constant at all.
Which mistakes cost marks?
Using v² = u² + 2as or v = u + at. The acceleration changes every instant.
Computing F = P/v at one speed and treating it as constant for the whole motion.
Using P = Fv with the net force when resistance acts. P = Fv applies to the driving force alone.
Mixing kW and W. Convert first: 20 kW = 20,000 W.
A practice routine
Derive v = √(2Pt/m) and x = mv³/(3P) from a blank page, twice this week.
Solve five ratio questions using only the proportionalities above.
Do two resistance problems: one asking for top speed, one asking for acceleration at a given speed.
Re-solve worked example 1 with the mass doubled. Predict first: speed should fall by a factor of √2, to about 14.1 m/s.
If you set up P = mv dv/dx and get stuck on the integral, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Tonight, derive x = mv³/3P without notes and check it against the 20 kW car.
