Let s be the required distance.

From Work - Energy theorem,

Work = Change in kinetic energy

$$\Rightarrow$$ Power $$\times$$ Time = $$\Delta$$K

i.e., Pt = $$\Delta$$K $$\Rightarrow$$ Pt = $${1 \over 2}$$mv² ..... (i)

Given, P = 1 Js^$$-$$1, t = 9 s, m = 2 kg

Substituting all the given values in eq. (i), we get

1 $$\times$$ 9 = $${1 \over 2}$$(2) v²

v² = 9 $$\Rightarrow$$ v = 3 m/s (at t = 9 s)

As, Fv = P $$\Rightarrow$$ (ma)v = P [$$\because$$ F = ma]

$$ \Rightarrow m\left[ {{{dv} \over {dt}}} \right]v = P \Rightarrow m\left[ {{{ds} \over {dt}}{{dv} \over {ds}}} \right]v = P$$

$$ \Rightarrow m\left[ {v{{dv} \over {ds}}} \right]v = P$$

$$ \Rightarrow 2{v^2}dv = ds$$ {$$\because$$ P = 1 J/s and m = 2 kg}

Integrating both sides,

$$\int\limits_0^3 {2{v^2}dv = \int\limits_0^s {ds \Rightarrow {2 \over 3}[{v^3}]_0^3 = 8} } $$

$${2 \over 3}[27 - 0] = s \Rightarrow s = 18$$ m

Hence, after 9 s, the body has moved a distance of 18 m.