Question

The engine of a car of mass m supplies constant power P to the wheel to accelerate the car. Rolling friction and air resistance can be neglected. The car is initially at rest. show that the displacement as a function of time is given by (x-xo) = (8p/9m)^1/2 . t^3/2

Answer 1

(x - x_0) = $\left(\frac{8P}{9m}\right)^{1/2} t^{3/2}$

Answer 2

1. The power ($P$) supplied by the engine is related to the force ($F$) and velocity ($v$) by $P = Fv$.
2. Using Newton's second law, $F = ma$, we substitute $F$ to get $P = mav$.
3. Since acceleration is the rate of change of velocity, $a = \frac{dv}{dt}$. Substituting this into the power equation gives $P = m \frac{dv}{dt} v$.
4. Rearranging the equation to separate variables: $v \, dv = \frac{P}{m} \, dt$.
5. Integrating both sides from initial conditions (car at rest, $v=0$ at $t=0$) to a general state ($v$ at time $t$): $\int_0^v v' \, dv' = \int_0^t \frac{P}{m} \, dt'$ $\frac{v^2}{2} = \frac{P}{m} t$ $v = \sqrt{\frac{2Pt}{m}}$
6. Velocity is the rate of change of displacement ($x$): $v = \frac{dx}{dt}$.
7. Substituting the expression for $v$: $\frac{dx}{dt} = \sqrt{\frac{2Pt}{m}}$.
8. Rearranging to separate variables: $dx = \sqrt{\frac{2P}{m}} \sqrt{t} \, dt$.
9. Integrating both sides from initial conditions (displacement $x_0$ at $t=0$) to a general state ($x$ at time $t$): $\int_{x_0}^x dx = \int_0^t \sqrt{\frac{2P}{m}} t'^{1/2} \, dt'$ $x - x_0 = \sqrt{\frac{2P}{m}} \left[ \frac{t'^{3/2}}{3/2} \right]_0^t$ $x - x_0 = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2}$
10. To match the required form, bring the constant $\frac{2}{3}$ inside the square root: $x - x_0 = \sqrt{\left(\frac{2}{3}\right)^2 \cdot \frac{2P}{m}} t^{3/2}$ $x - x_0 = \sqrt{\frac{4}{9} \cdot \frac{2P}{m}} t^{3/2}$ $x - x_0 = \sqrt{\frac{8P}{9m}} t^{3/2}$