Question

An automobile of mass 'm' accelerates starting from origin and initially at rest, while the engine supplies constant power P. The position is given as a function of time by:

• A. $(\frac{9P}{8m})^{1/3} t^{2}$
• B. $(\frac{8P}{9m})^{1/2} t^{2/3}$
• C. $(\frac{8P}{9m})^{1/2} t^{3/2}$
• D. $(\frac{9m}{8P})^{1/2} t^{3/2}$

Hint:

For constant power problems, use the work-energy relationship $P = dK/dt$. Integrating this gives $K(t)$, from which you can find $v(t)$. A second integration of $v(t)$ yields the position $x(t)$.

Answer 1

The Correct Option is C

When an engine supplies constant power P, this power is used to increase the kinetic energy of the automobile.
Power $P = \frac{dK}{dt}$, where K is the kinetic energy.
Since the automobile starts from rest, its kinetic energy at any time t is $K = \frac{1}{2}mv^2$.
$P = \frac{d}{dt} \left(\frac{1}{2}mv^2\right)$.
Since P is constant, we can integrate with respect to time to find the kinetic energy:
$\int_0^t P dt = \int_0^K dK \implies Pt = K = \frac{1}{2}mv^2$.
From this, we can find the velocity v as a function of time.
$v^2 = \frac{2Pt}{m} \implies v = \sqrt{\frac{2Pt}{m}} = \left(\frac{2P}{m}\right)^{1/2} t^{1/2}$.
Position x is the integral of velocity v with respect to time. Since it starts from the origin, $x(0)=0$.
$x = \int_0^t v dt = \int_0^t \left(\frac{2P}{m}\right)^{1/2} t^{1/2} dt$.
$x = \left(\frac{2P}{m}\right)^{1/2} \int_0^t t^{1/2} dt$.
$x = \left(\frac{2P}{m}\right)^{1/2} \left[ \frac{t^{3/2}}{3/2} \right]_0^t = \left(\frac{2P}{m}\right)^{1/2} \frac{2}{3} t^{3/2}$.
To match the format of the options, we bring the constant $\frac{2}{3}$ inside the square root.
$x = \sqrt{\left(\frac{2}{3}\right)^2 \left(\frac{2P}{m}\right)} \ t^{3/2}$.
$x = \sqrt{\frac{4}{9} \cdot \frac{2P}{m}} \ t^{3/2} = \sqrt{\frac{8P}{9m}} \ t^{3/2}$.
This can be written as $x = \left(\frac{8P}{9m}\right)^{1/2} t^{3/2}$.