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(C) For statement $A$: The kinetic energy $KE$ of a body starting from rest under a constant force $F$ is given by $KE = \frac{1}{2}mv^2$. Since $v =

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$A$ is correct and $B$ is wrong

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(C) For statement $A$: The kinetic energy $KE$ of a body starting from rest under a constant force $F$ is given by $KE = \frac{1}{2}mv^2$. Since $v = at$,we have $KE = \frac{1}{2}m(at)^2 = \frac{1}{2}ma^2t^2$.The rate of change of kinetic energy is $\frac{d(KE)}{dt} = \frac{d}{dt}(\frac{1}{2}ma^2t^2) = ma^2t$.Since $m$ and $a$ are constants,$\frac{d(KE)}{dt} \propto t$. Thus,the rate of change of kinetic energy varies linearly with time. Statement $A$ is correct.For statement $B$: $A$ body at rest is in equilibrium only if the net force acting on it is zero. $A$ body can be momentarily at rest (e.g.,a ball thrown vertically upward at its highest point) while being acted upon by a non-zero net force (gravity). Thus,it is not necessarily in equilibrium. Statement $B$ is wrong.

Consider the following statements $A$ and $B$ and identify the correct answer given below:
$(A)$ $A$ body initially at rest is acted upon by a constant force. The rate of change of its kinetic energy varies linearly with time.
$(B)$ When a body is at rest,it must be in equilibrium.

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Consider the following statements $A$ and $B$ and identify the correct answer given below:$(A)$ $A$ body initially at rest is acted upon by a constant force. The rate of change of its kinetic energy varies linearly with time.$(B)$ When a body is at rest,it must be in equilibrium.