Power of a Vehicle Engine

The situation

A vehicle of mass $m$ moves at velocity $v$. Its engine exerts a driving force $F$ at the wheels. Road resistance (friction, air drag) totals $f_r$.

Net force: $F_{net} = F - f_r = ma$

Driving force: $F = ma + f_r$

Power delivered by engine:

$P = Fv = (ma + f_r)v$

Three operating conditions

Accelerating ($a > 0$):

$P = (ma + f_r)v$

More power needed at higher speeds (both terms increase with $v$ indirectly — $v$ increases while $a$ and $f_r$ may change).

Constant speed ($a = 0$):

$P = f_r \cdot v$

All engine power goes to overcome resistance. No power is needed for acceleration.

Maximum speed:

At maximum speed, engine power equals resistance power:

$P_{max} = f_r \cdot v_{max}$

$v_{max} = \frac{P_{max}}{f_r}$

This is the terminal velocity for a vehicle — the speed at which engine power exactly balances resistance.

Power vs speed relationship

At constant power $P$:

$F = \frac{P}{v}$

As speed increases, the available driving force decreases (inverse relationship). The acceleration:

$a = \frac{F - f_r}{m} = \frac{P/v - f_r}{m}$

As $v$ increases, $a$ decreases. Eventually $a = 0$ at $v_{max}$.

Example

A car of mass 1000 kg has engine power 50 kW and road resistance 500 N. Find maximum speed and acceleration at 20 m/s.

Maximum speed:

$v_{max} = \frac{P}{f_r} = \frac{50000}{500} = 100 \text{ m/s} \approx 360 \text{ km/h}$

(This is unrealistically high — real cars have much higher resistance at high speeds due to aerodynamic drag increasing as $v^2$.)

Acceleration at $v = 20$ m/s:

$F = \frac{P}{v} = \frac{50000}{20} = 2500 \text{ N}$

$a = \frac{F - f_r}{m} = \frac{2500 - 500}{1000} = 2 \text{ m/s}^2$