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(B) Power $P$ is given by $P = Fv = (ma)v = m(\frac{dv}{dt})v$. Since $P$ is constant,$m v dv = P dt$. Integrating both sides,$\int_0^v m v dv = \int_

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Power $P$ is given by $P = Fv = (ma)v = m(\frac{dv}{dt})v$. Since $P$ is constant,$m v dv = P dt$. Integrating both sides,$\int_0^v m v dv = \int_0^t P dt$,which gives $\frac{1}{2} m v^2 = Pt$. So,$v^2 = \frac{2Pt}{m}$,or $v = (\frac{2P}{m})^{1/2} t^{1/2}$. Since $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} = (\frac{2P}{m})^{1/2} t^{1/2}$. Integrating with respect to $t$,$s = \int_0^t (\frac{2P}{m})^{1/2} t^{1/2} dt$. $s = (\frac{2P}{m})^{1/2} [\frac{t^{3/2}}{3/2}] = (\frac{2P}{m})^{1/2} \cdot \frac{2}{3} t^{3/2}$. $s = (\frac{4}{9} \cdot \frac{2P}{m})^{1/2} t^{3/2} = (\frac{8P}{9m})^{1/2} t^{3/2}$.

$A$ vehicle of mass $m$ starts from rest and accelerates. If the engine delivers a constant power $P$,then the position $s$ of the vehicle at time $t$ is given by:

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