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(A) Given the equation of motion: $s = \frac{1}{2}vt$. Multiplying by $2$,we get $2s = vt$. Differentiating both sides with respect to $t$: $2 \frac{d

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Constant

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(A) Given the equation of motion: $s = \frac{1}{2}vt$. Multiplying by $2$,we get $2s = vt$. Differentiating both sides with respect to $t$: $2 \frac{ds}{dt} = v + t \frac{dv}{dt}$. Since $v = \frac{ds}{dt}$,we substitute this into the equation: $2v = v + t \frac{dv}{dt} \implies v = t \frac{dv}{dt}$. Since acceleration $a = \frac{dv}{dt}$,we have $v = ta$. Differentiating again with respect to $t$: $\frac{dv}{dt} = a + t \frac{da}{dt}$. Substituting $a = \frac{dv}{dt}$: $a = a + t \frac{da}{dt} \implies t \frac{da}{dt} = 0$. Since $t 
eq 0$ for motion,we must have $\frac{da}{dt} = 0$. This implies that the acceleration $a$ is constant.

If the law of motion in a straight line is $s = \frac{1}{2}vt$,then acceleration is

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