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(C) For statement $A$:Kinetic energy $KE = \frac{1}{2} m v^2$. Since the body starts from rest and is acted upon by a constant force $F$, its accelera

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$A$ is correct and $B$ is wrong

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(C) For statement $A$:Kinetic energy $KE = \frac{1}{2} m v^2$. Since the body starts from rest and is acted upon by a constant force $F$, its acceleration $a = F/m$ is constant.Thus, $v = at$, and $KE = \frac{1}{2} m (at)^2 = \frac{1}{2} m a^2 t^2$.The rate of change of kinetic energy is $\frac{d(KE)}{dt} = \frac{d}{dt} (\frac{1}{2} m a^2 t^2) = m a^2 t$.Since $\frac{d(KE)}{dt} \propto t$, the rate of change of kinetic energy varies linearly with time. Thus, 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 still being acted upon by a non-zero net force (gravity). Therefore, it is not necessarily in equilibrium. Thus, 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.