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(A) Given the equation: $t = \sqrt{x} + 2$.Rearranging for $x$: $\sqrt{x} = t - 2$,which implies $x = (t - 2)^2$.The velocity $v$ is the derivative of

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(A) Given the equation: $t = \sqrt{x} + 2$.Rearranging for $x$: $\sqrt{x} = t - 2$,which implies $x = (t - 2)^2$.The velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = 2(t - 2)$.At $t = 0 \ s$,the initial velocity $u = v(0) = 2(0 - 2) = -4 \ m/s$.At $t = 4 \ s$,the final velocity $v_f = v(4) = 2(4 - 2) = 4 \ m/s$.According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m u^2$.Substituting the values: $W = \frac{1}{2} m (4)^2 - \frac{1}{2} m (-4)^2 = \frac{1}{2} m (16) - \frac{1}{2} m (16) = 0 \ J$.

The position $x$ of a particle moving along the $x$-axis at time $t$ is given by the equation $t = \sqrt{x} + 2$,where $x$ is in metres and $t$ is in seconds. The work done by the force in the first $4 \ s$ is .............. $J$.

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