Kvant Physics Problem 16

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Problem

A car of weight $P$, with both axles driven, starts from rest. The car’s engine operates at constant power $W$, and the coefficient of sliding friction between the wheels and the road is $k$. Find the dependence of the car’s velocity on time and sketch the graph of this dependence. Neglect air resistance and friction in the mechanisms.

A. T. Drozdov

Setup and Assumptions

A car moves on a horizontal road from rest under the action of an engine delivering constant power $W$ in watts. The car’s weight is $P$ in newtons, so its mass is $m=\frac{P}{g}$ in kilograms, where $g$ is the gravitational acceleration in $\mathrm{m/s^2}$. The coefficient of sliding friction between tires and road is $k$, so the maximal tractive force that can be transmitted without slipping is $F_{\max}=kP$ in newtons.

The motion is one-dimensional along a straight road in an inertial reference frame attached to the ground. Air resistance, drivetrain losses, and rotational inertia of wheels are neglected. The engine is assumed to be capable of delivering constant mechanical power to the wheels. Both axles are driven, so the total available friction force is $kP$.

The unknown is the velocity $v(t)$ of the car as a function of time.

Physical Principles

Newton’s second law for translational motion is

$m\frac{dv}{dt}=F.$

Mechanical power delivered by the engine is

$W = Fv,$

so the tractive force provided by the engine is

$F_{\text{eng}}=\frac{W}{v}.$

The actual traction force is limited by static friction,

$F \le kP.$

Thus the realized force is

$F=\min!\left(kP,\frac{W}{v}\right).$

Derivation

At the initial stage the velocity is small, so the expression $\frac{W}{v}$ exceeds the friction limit $kP$, since it diverges as $v \to 0$. The motion is therefore friction-limited, and the traction force equals

$F=kP.$

Newton’s second law gives

$m\frac{dv}{dt}=kP.$

Substituting $m=\frac{P}{g}$ yields

$\frac{P}{g}\frac{dv}{dt}=kP,$

$\frac{dv}{dt}=kg.$

Integration with initial condition $v(0)=0$ gives

$v(t)=kgt.$

The transition to the power-limited regime occurs when

$\frac{W}{v}=kP,$

which defines the critical velocity

$v_*=\frac{W}{kP}.$

The corresponding transition time $t__$ follows from $v(t__)=v_*$:

$kgt_*=\frac{W}{kP},$

$t_*=\frac{W}{k^2Pg}.$

For $t \ge t_*$ the motion is power-limited, so the force is

$F=\frac{W}{v}.$

Newton’s second law becomes

$m\frac{dv}{dt}=\frac{W}{v}.$

Substituting $m=\frac{P}{g}$ gives

$\frac{P}{g}\frac{dv}{dt}=\frac{W}{v},$

hence

$v\frac{dv}{dt}=\frac{Wg}{P}.$

Rewriting,

$v,dv=\frac{Wg}{P},dt.$

Integration from $t_*$ to $t$ yields

$\frac{1}{2}\left(v^2-v_^2\right)=\frac{Wg}{P}(t-t_).$

Solving for $v(t)$ gives

$v(t)=\sqrt{v_^2+\frac{2Wg}{P}(t-t_)}.$

Result

The velocity is

$$v(t)= \begin{cases} kgt, & 0 \le t \le \frac{W}{k^2Pg},\[6pt] \sqrt{\left(\frac{W}{kP}\right)^2+\frac{2Wg}{P}\left(t-\frac{W}{k^2Pg}\right)}, & t \ge \frac{W}{k^2Pg}. \end{cases}$$

The first regime corresponds to linear growth with slope $kg$, and the second regime corresponds to a square-root dependence after the transition.

Sanity Checks

Dimensional consistency holds since $kg$ has units $\mathrm{m/s^2}$, giving $v=kg t$ in $\mathrm{m/s}$, and the second term under the square root has units $\mathrm{(m/s)^2}$ because $\frac{Wg}{P}$ has units $\mathrm{m^2/s^3}$ multiplied by time.

At $t=0$, the velocity is zero, consistent with the initial condition. At $t=t__$ both expressions give $v=v__=\frac{W}{kP}$, ensuring continuity of velocity at the regime change.

For very small $W$, the transition time $t_*=\frac{W}{k^2Pg}$ becomes small and the motion quickly enters the power-limited regime. For very large $W$, the friction-limited regime persists longer, and the initial linear growth dominates.

The acceleration in the first stage is constant $kg$, so any sign error would most likely appear when substituting $m=\frac{P}{g}$ or when equating the friction force to $kP$.