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(A) Power $P$ is given by $P = Fv = (ma)v = m \left(\frac{dv}{dt}\right)v$.Since $P$ is constant,$P = mv \frac{dv}{dt}$.Integrating with respect to ti

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Power $P$ is given by $P = Fv = (ma)v = m \left(\frac{dv}{dt}\right)v$.Since $P$ is constant,$P = mv \frac{dv}{dt}$.Integrating with respect to time: $\int_{0}^{v} mv \, dv = \int_{0}^{t} P \, dt$.$\frac{1}{2}mv^2 = Pt \implies v = \sqrt{\frac{2Pt}{m}} = \left(\frac{2P}{m}\right)^{1/2} t^{1/2}$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \left(\frac{2P}{m}\right)^{1/2} t^{1/2}$.Integrating with respect to position and time: $\int_{0}^{x} dx = \int_{0}^{t} \left(\frac{2P}{m}\right)^{1/2} t^{1/2} dt$.$x = \left(\frac{2P}{m}\right)^{1/2} \left[ \frac{t^{3/2}}{3/2} \right] = \left(\frac{2P}{m}\right)^{1/2} \cdot \frac{2}{3} t^{3/2}$.$x = \left( \frac{4}{9} \cdot \frac{2P}{m} \right)^{1/2} t^{3/2} = \left( \frac{8P}{9m} \right)^{1/2} t^{3/2}$.

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

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