A body moving with initial velocity u accelerates until it attains velocity v within time t. The distance covered is given by which expression?

Step 1: Identify the given values. The body starts from rest, so the initial velocity is $u = 0 ms^{-1}$. The acceleration is $a = 2 ms^{-2}$. The distance travelled is $s = 9 \text{ m}$. We need to find the final velocity $v$.

Step 2: Choose the appropriate kinematic equation. The kinematic equation that relates initial velocity ($u$), final velocity ($v$), acceleration ($a$), and displacement ($s$) is: $v^2 = u^2 + 2as$

Step 3: Substitute the given values into the equation. $v^2 = (0 ms^{-1})^2 + 2(2 ms^{-2})(9 m)$ $v^2 = 0 + 36 m^2s^{-2}$ $v^2 = 36 m^2s^{-2}$

Step 4: Solve for $v$. $v = \sqrt{36 m^2s^{-2}}$ $v = 6 ms^{-1}$

The magnitude of its velocity after travelling $9 \text{ m}$ is $\boxed{6.0 ms^{-1}}$.