Kvant Physics Problem 16
The physical system is a car moving along a horizontal road.
Verified: yes
Verdicts: PASS + PASS
Solve time: 11m19s
Source on kvant.digital
Problem
A car of weight $P$, with both axles driven, starts from rest. The car’s engine operates at a constant power $W$, and the coefficient of sliding friction between the wheels and the road is $k$. Find the dependence of the car’s speed on time and draw the graph of this dependence. Neglect air resistance and friction in the mechanisms.
A. T. Drozdov
Setup and Assumptions
The physical system is a car moving along a horizontal road. The car starts from rest and has both axles driven. Its weight is $P$, measured in newtons. The engine develops a constant power $W$, measured in watts. The coefficient of sliding friction between the wheels and the road is $k$.
The unknown quantity is the speed $v(t)$ of the car as a function of time.
The motion is considered in an inertial reference frame fixed to the Earth. Air resistance is neglected. Mechanical losses in the transmission and engine are neglected. The road is horizontal, so the normal reaction of the road equals the weight of the car.
The driving force transmitted to the road cannot exceed the friction force. Since both axles are driven, the maximum possible traction force is
$$F_{\max}=kP.$$
The mass of the car is
$$m=\frac{P}{g},$$
where $g$ is the acceleration due to gravity.
Physical Principles
The solution is based on Newton's second law,
$$m\frac{dv}{dt}=F,$$
where $F$ is the driving force acting on the car.
The engine power is constant:
$$W=Fv.$$
Consequently, if the engine is not limited by wheel slip, the driving force is
$$F=\frac{W}{v}.$$
The traction force cannot exceed the maximum friction force,
$$F\le kP.$$
Hence the actual driving force is determined by the smaller of the two quantities,
$$F=\min!\left(kP,\frac{W}{v}\right).$$
Derivation
At very small speeds, the quantity $W/v$ is very large. The condition
$$\frac{W}{v}>kP$$
holds, so the available engine power would require a traction force greater than friction can provide. The wheels slip, and the actual force equals the friction limit:
$$F=kP.$$
Newton's second law gives
$$m\frac{dv}{dt}=kP.$$
Substituting $m=P/g$,
$$\frac{P}{g}\frac{dv}{dt}=kP,$$
or
$$\frac{dv}{dt}=kg.$$
Integrating from $t=0$, when $v=0$,
$$v=kg,t.$$
This regime continues until the force required by the constant-power engine becomes equal to the maximum friction force. The transition speed $v_1$ is found from
$$\frac{W}{v_1}=kP.$$
Hence
$$v_1=\frac{W}{kP}.$$
The corresponding transition time is
$$t_1=\frac{v_1}{kg} =\frac{W}{k^2Pg}.$$
For $t>t_1$, wheel slip no longer limits the motion. The engine power condition determines the force:
$$F=\frac{W}{v}.$$
Newton's second law becomes
$$m\frac{dv}{dt}=\frac{W}{v}.$$
Substituting $m=P/g$,
$$\frac{P}{g}\frac{dv}{dt}=\frac{W}{v}.$$
Multiplying by $v$,
$$v\frac{dv}{dt}=\frac{Wg}{P}.$$
Since
$$v\frac{dv}{dt} = \frac12\frac{d(v^2)}{dt},$$
we obtain
$$\frac12\frac{d(v^2)}{dt} = \frac{Wg}{P}.$$
Integrating from the transition instant $(t_1,v_1)$,
$$v^2-v_1^2 = \frac{2Wg}{P}(t-t_1).$$
Thus
$$v= \sqrt{ v_1^2+\frac{2Wg}{P}(t-t_1) }.$$
Substituting
$$v_1=\frac{W}{kP}, \qquad t_1=\frac{W}{k^2Pg},$$
gives
$$v= \sqrt{ \left(\frac{W}{kP}\right)^2 + \frac{2Wg}{P} \left( t-\frac{W}{k^2Pg} \right) }.$$
After simplification,
$$v= \sqrt{ \frac{2Wg}{P},t - \left(\frac{W}{kP}\right)^2 }.$$
This expression is valid only for $t\ge t_1$.
Result
The speed of the car is
$$v(t)= \begin{cases} kg,t, & 0\le t\le \dfrac{W}{k^2Pg}, \[2ex] \sqrt{ \dfrac{2Wg}{P},t - \left(\dfrac{W}{kP}\right)^2 }, & t\ge \dfrac{W}{k^2Pg}. \end{cases}$$
The transition occurs at
$$t_1=\frac{W}{k^2Pg}, \qquad v_1=\frac{W}{kP}.$$
No numerical values are provided in the statement, so a numerical substitution cannot be performed.
The graph consists of two smoothly joined parts. From $t=0$ to $t=t_1$, the speed increases linearly with slope $kg$. After $t=t_1$, the speed follows a square-root law. The curve continues to rise, but its slope decreases with time.
Sanity Checks
The dimensions are correct. In the first regime,
$$kg,t$$
has dimensions
$$(1)\left(\frac{\text{m}}{\text{s}^2}\right)\text{s} = \frac{\text{m}}{\text{s}},$$
which is a speed.
In the second regime,
$$\frac{Wg}{P} = \frac{\left(\text{N}\cdot\text{m}/\text{s}\right) (\text{m}/\text{s}^2)} {\text{N}} = \frac{\text{m}^2}{\text{s}^3},$$
and multiplying by $t$ gives $\text{m}^2/\text{s}^2$, the correct dimension for $v^2$.
If $k\to\infty$, there is no traction limitation. Then
$$t_1\to0,$$
and the solution reduces immediately to
$$v=\sqrt{\frac{2Wg}{P},t},$$
which is the standard result for motion under constant power.
If the engine power is very small, $W\to0$, then both $v_1$ and the subsequent acceleration tend to zero, consistent with physical expectations.
The most delicate step is the transition between the two regimes. An incorrect substitution of
$$v_1=\frac{W}{kP}$$
or
$$t_1=\frac{W}{k^2Pg}$$
into the integrated constant-power solution would destroy continuity of the speed at the joining point. The obtained expressions satisfy
$$v(t_1)=\frac{W}{kP},$$
from both sides, confirming the correctness of the result.