Question 103 of 308

A particle is moving unidirectional on a horizontal plane under the action of a constant power supplying energy source. The displacement (s) – time (t) graph that describes the motion of the particle is (graphs are drawn schematically and are not to scale):

ABCD

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Answer: (B)

Here Power is constant,
Hence, $P=Fv=mav=av=constant.$
$av=k v\dfrac{dv}{dx}\times v=k {\int }_{0}^{v}{v}^{2}dv=k{\int }_{0}^{x}dx ∴v={\left(3k\right)}^{1/3}\times {\left(x\right)}^{1/3}\Rightarrow v=A{\left(x\right)}^{1/3}whereA={\left(3k\right)}^{1/3}=constant....\left(1\right)$
Also, $v=\dfrac{dx}{dt}$$∴\dfrac{dx}{dt}=A{\left(x\right)}^{1/3}\Rightarrow {\int }_{0}^{d}\dfrac{dx}{{\left(x\right)}^{1/3}}=A{\int }_{0}^{t}dt$
$Simplifying further we get,⇒d=At3/2 ...2$
From equations \left(1\right)&\left(2\right) its clear that the graph won't be straight line so first & third options cant be correct.
Now, from equation $\left(1\right)$ its clear that when $x$ is increased then the slope $\dfrac{ds}{dt}$ should also increase which is possible in second option. Hence option $2$ is correct.