A particle is moving unidirectional on a horizontal plane under the...

Here Power is constant,
Hence, $P = Fv = mav = av = constant .$

$av = k$
$v\dfrac{dv}{dx} \times v = k$
${∫}_{0}^{v}{v}^{2}dv = k{∫}_{0}^{x}dx$
$∴ v = {\left(3k\right)}^{1 / 3} \times {\left(x\right)}^{1 / 3} ⇒ v = A{\left(x\right)}^{1 / 3}whereA = {\left(3k\right)}^{1 / 3} = constant . . . . \left(1\right)$

Also, $v = \dfrac{dx}{dt}$$∴ \dfrac{dx}{dt} = A{\left(x\right)}^{1 / 3} ⇒ {∫}_{0}^{d}\dfrac{dx}{{\left(x\right)}^{1 / 3}} = A{∫}_{0}^{t}dt$

Simplifying further we get,

$⇒ d = A{\left(t\right)}^{3 / 2} . . . \left(2\right)$

From equations \left(1\right) \text{ and } \left(2\right)its clear that the graph won't be straight line so first & third options cant be correct.<br/>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.