A body of mass 2, kg is driven by an engine d | vedclass

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(D) Given: Mass $m = 2\, kg$,Power $P = 1\, J/s$,Time $t = 9\, s$,Initial velocity $u = 0$.Power is defined as $P = F \cdot v = (m \cdot a) \cdot v =

Vedclass · Accepted Answer

$18$

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(D) Given: Mass $m = 2\, kg$,Power $P = 1\, J/s$,Time $t = 9\, s$,Initial velocity $u = 0$.Power is defined as $P = F \cdot v = (m \cdot a) \cdot v = m \cdot \frac{dv}{dt} \cdot v$.Rearranging the terms: $m \cdot v \cdot dv = P \cdot dt$.Integrating both sides from rest ($v=0$ at $t=0$): $\int_{0}^{v} m \cdot v \cdot dv = \int_{0}^{t} P \cdot dt$.$m \cdot \frac{v^2}{2} = P \cdot t \Rightarrow v^2 = \frac{2Pt}{m} \Rightarrow v = \sqrt{\frac{2Pt}{m}}$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \sqrt{\frac{2P}{m}} \cdot t^{1/2}$.Integrating to find distance $x$: $\int_{0}^{x} dx = \sqrt{\frac{2P}{m}} \int_{0}^{t} t^{1/2} dt$.$x = \sqrt{\frac{2P}{m}} \cdot \frac{t^{3/2}}{3/2} = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} \cdot t^{3/2}$.Substituting the values $m = 2$,$P = 1$,$t = 9$: $x = \sqrt{\frac{2 \cdot 1}{2}} \cdot \frac{2}{3} \cdot 9^{3/2} = 1 \cdot \frac{2}{3} \cdot (3^2)^{3/2} = \frac{2}{3} \cdot 27 = 18\, m$.

$A$ body of mass $2\, kg$ is driven by an engine delivering a constant power $1\, J/s$. The body starts from rest and moves in a straight line. After $9$ seconds,the body has moved a distance (in $m$):

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