AP® Physics C: Mechanics · Unit 3 · Topic 3.5

Instantaneous Power

Instantaneous power is the rate at which force does work at a single moment, given by P = F dot v. It tells you how hard an engine is working right now, not over a whole trip.

Work is force acting along a path, and power is just how fast that work gets done. Start from W as the integral of F dot dr, and take the time derivative: dW/dt = F dot (dr/dt) = F dot v. That's instantaneous power. When force and velocity point the same way, the dot product collapses to a plain product, P = Fv. This is why a car's engine strains hardest at highway speed rather than at a stoplight: the driving force needed to fight resistance barely changes, but velocity has climbed, so the product Fv, the power demand, climbs with it.

This single equation also explains top speed. A real engine has a maximum power output it can sustain, not a maximum force. As a car speeds up, resistive forces (drag, friction) stay roughly fixed or grow, but the velocity needed to turn a given force into the engine's full power output keeps rising. Top speed is reached when all available power is spent just cancelling resistance at constant velocity: P = f v, with zero leftover power to accelerate. Solve for v and you have the ceiling.

Power connects directly to the energy accounting of this unit. If you know P(t), integrating it over time hands you back total work done, W = integral of P dt, which is just the area under a power versus time graph. That's the same work that feeds into the work-energy theorem and potential energy bookkeeping covered elsewhere in this unit: force tells you what's pushing, power tells you how fast energy is changing hands, and work tells you the total transferred.

Key terms

Instantaneous power
The rate at which work is being done at a specific moment, P = dW/dt.
P = F dot v
Power as the dot product of the force vector and the velocity vector at an instant.
Top speed
The constant velocity reached when all of an engine's power output is used just to balance resistive force.
Area under a power-time graph
The integral of power over time, which equals the total work done.

Watch it explained

  1. How it worksHow does P equal to force dotted with velocity fall straight out of dW/dt?
  2. Worked exampleFind the instantaneous power delivered by a 300 N driving force to a car moving at 28 m/s.
  3. Worked exampleFind the top speed of a 150 kW car that faces a total resistive force of 900 N.
  4. Ties togetherHow does the area under a power versus time graph get you back to the total work?

Check yourself

Your turn1 of 2
Find the instantaneous power delivered by a 379 N driving force to a car moving at 29 m/s.

End of topic.