How does a car move when it starts at constant power?

With P fixed and P = Fv, the traction F = P/v drops as v rises, so the acceleration a = (F − F_r)/m keeps decreasing. The car speeds up with ever smaller acceleration, and its v–t graph is a curve that keeps flattening. When the traction has fallen to equal the resistance, a = 0 and the speed is at its maximum v_max = P/F_r; after that it cruises.

How long can the uniform acceleration last?

During uniform acceleration the traction is constant: F − F_r = ma, so F = F_r + ma. As speed rises the power P = Fv grows linearly, and the uniform phase ends when it reaches the rated power: v₁ = P_rated/(F_r + ma), after t₁ = v₁/a. Note that v₁ is less than v_max = P_rated/F_r; the car then continues at rated power with shrinking acceleration and only later reaches v_max.

Why can’t SUVAT give the distance in the constant-power phase?

In that phase the acceleration changes all the time, so it is not uniformly accelerated motion and formulas such as x = v₀t + ½at² do not apply. The traction’s work is simply W = Pt and the resistance does −F_r·s, so the work–energy theorem Pt − F_r·s = ½mv² − ½mv₀² gives the distance. The tool shows this substitution and compares it with the simulated distance; they agree.

What is the top speed going uphill?

Uphill the car must overcome both the resistance f and the downhill component of gravity mg sinθ. At top speed the traction equals their sum, F = f + mg sinθ, so v_max = P/(f + mg sinθ), lower than on the level. In a constant-acceleration start the traction during the uniform phase becomes F = f + mg sinθ + ma.

What happens if the power is suddenly halved at cruising speed?

The speed cannot change instantly, so by P = Fv the traction immediately halves and becomes smaller than the resistance, and the car slows down. As it slows, F = P′/v grows and the deceleration shrinks, until traction again equals resistance and the car cruises at the new top speed v′ = P′/F_r, half the old one. The preset “Power halved at top speed” shows this.

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🚗Vehicle Start-up - Constant Power vs Constant Acceleration, v–t Graphs, Work–Energy

This tool simulates the two ways a car can start in exam problems. At constant power the traction F = P/v falls as speed rises, so the acceleration keeps shrinking until v_max = P/F_r. At constant acceleration the car first speeds up uniformly while its power rises linearly, then switches to constant power once the rating is reached. It plots v–t, a–t, P–t and F–t together, animates the car with traction and resistance arrows, explains every phase with the numbers substituted, and finds the constant-power distance from the work–energy theorem, checked against the simulation. Slopes and a sudden power change are supported.

Free · No sign-up · Runs in your browserLast updated: 2026-10-29

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How to use Vehicle Start-up (Power vs Acceleration)

1Choose constant acceleration or constant power, or pick a classic case
2Set the mass, rated power, resistance, slope and acceleration, and optionally a time when the power changes
3Play the animation and read the four graphs and the phase analysis; drag the time bar to any moment
4Open Compare to see both modes for the same car, or Practice for random problems

FAQ

How does a car move when it starts at constant power?: With P fixed and P = Fv, the traction F = P/v drops as v rises, so the acceleration a = (F − F_r)/m keeps decreasing. The car speeds up with ever smaller acceleration, and its v–t graph is a curve that keeps flattening. When the traction has fallen to equal the resistance, a = 0 and the speed is at its maximum v_max = P/F_r; after that it cruises.
How long can the uniform acceleration last?: During uniform acceleration the traction is constant: F − F_r = ma, so F = F_r + ma. As speed rises the power P = Fv grows linearly, and the uniform phase ends when it reaches the rated power: v₁ = P_rated/(F_r + ma), after t₁ = v₁/a. Note that v₁ is less than v_max = P_rated/F_r; the car then continues at rated power with shrinking acceleration and only later reaches v_max.
Why can’t SUVAT give the distance in the constant-power phase?: In that phase the acceleration changes all the time, so it is not uniformly accelerated motion and formulas such as x = v₀t + ½at² do not apply. The traction’s work is simply W = Pt and the resistance does −F_r·s, so the work–energy theorem Pt − F_r·s = ½mv² − ½mv₀² gives the distance. The tool shows this substitution and compares it with the simulated distance; they agree.
What is the top speed going uphill?: Uphill the car must overcome both the resistance f and the downhill component of gravity mg sinθ. At top speed the traction equals their sum, F = f + mg sinθ, so v_max = P/(f + mg sinθ), lower than on the level. In a constant-acceleration start the traction during the uniform phase becomes F = f + mg sinθ + ma.
What happens if the power is suddenly halved at cruising speed?: The speed cannot change instantly, so by P = Fv the traction immediately halves and becomes smaller than the resistance, and the car slows down. As it slows, F = P′/v grows and the deceleration shrinks, until traction again equals resistance and the car cruises at the new top speed v′ = P′/F_r, half the old one. The preset “Power halved at top speed” shows this.

Features

Full simulation of both start-up modes, with the constant-power phase integrated numerically
v–t, a–t, P–t and F–t graphs in sync, marking v₁, v_max, rated power and resistance
Calculates v_max = P/F_r, v₁ = P/(F_r + ma), t₁ and s₁
Constant-power distance from Pt − F_r·s = ½m(v² − v₀²), compared with the simulated value
Uphill driving (resistance plus mg sinθ) and a sudden power change mid-drive
Side-by-side v–t comparison of both modes and random practice on top speed, t₁ and distance

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