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

We are told that the engine supplies a constant power $$P = 1\,\text{J s}^{-1}$$ to a body of mass $$m = 2\,\text{kg}$$. The body starts from rest, so its initial velocity is $$v_0 = 0$$. Power is defined as the rate at which work is done, and, for translational motion in a straight line, we have the relation

$$P \;=\; \frac{dW}{dt} \;=\; F\,v$$

Because the force responsible for the motion provides the acceleration, we can rewrite the force in terms of mass and acceleration: $$F = m\,a$$. Substituting this into the power expression gives

$$P \;=\; (m\,a)\,v \;=\; m\,a\,v$$

The acceleration $$a$$ is the time derivative of velocity, $$a = \dfrac{dv}{dt}$$. Replacing $$a$$ by $$\dfrac{dv}{dt}$$, we obtain

$$P \;=\; m\,v\,\frac{dv}{dt}$$

All the quantities in this differential equation are separable, so we rearrange to isolate the variables:

$$v\,dv \;=\; \frac{P}{m}\,dt$$

Now we integrate both sides from the initial state (velocity $$0$$ at time $$0$$) to some general state (velocity $$v$$ at time $$t$$):

$$\int_{0}^{v} v\,dv \;=\; \frac{P}{m}\int_{0}^{t} dt$$

Evaluating the integrals, we get

$$\frac{1}{2}v^{2} \;=\; \frac{P}{m}\,t$$

Multiplying by $$2$$ yields an explicit expression for the square of the velocity:

$$v^{2} \;=\; 2\,\frac{P}{m}\,t$$

Taking the positive square root (because the body moves forward), we find the velocity as a function of time:

$$v(t) \;=\; \sqrt{\frac{2P}{m}\,t}$$

The displacement $$s$$ in time $$t$$ is the time integral of the velocity:

$$s \;=\; \int_{0}^{t} v(t')\,dt' \;=\; \int_{0}^{t} \sqrt{\frac{2P}{m}\,t'}\,dt'$$

We factor out the constants to simplify the integral:

$$s \;=\; \sqrt{\frac{2P}{m}}\;\int_{0}^{t} (t')^{1/2}\,dt'$$

We now integrate the power-law function. The integral of $$t'^{1/2}$$ is $$\tfrac{2}{3}t'^{3/2}$$. Substituting the bounds gives

$$s \;=\; \sqrt{\frac{2P}{m}}\;\left[\frac{2}{3}\,t^{3/2}\right]$$

This can be rewritten more compactly as

$$s \;=\; \frac{2}{3}\,\sqrt{\frac{2P}{m}}\;t^{3/2}$$

We now substitute the numerical values $$P = 1\,\text{J s}^{-1}$$, $$m = 2\,\text{kg}$$ and $$t = 9\,\text{s}$$.

First, calculate the square-root factor:

$$\sqrt{\frac{2P}{m}} \;=\; \sqrt{\frac{2 \times 1}{2}} \;=\; \sqrt{1} \;=\; 1$$

With this, the expression for $$s$$ simplifies to

$$s \;=\; \frac{2}{3}\;t^{3/2}$$

Next, evaluate $$t^{3/2}$$ for $$t = 9\,\text{s}$$. We note that $$9^{1/2} = 3$$, so

$$9^{3/2} \;=\; (9^{1/2})^{3} \;=\; 3^{3} \;=\; 27$$

Substituting this value, we find

$$s \;=\; \frac{2}{3}\times 27 \;=\; 18$$

Therefore, after $$9\,\text{s}$$ the body has travelled a distance of $$18\,\text{m}$$.

So, the answer is $$18\,\text{m}$$.