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(B) Given that the distance $s$ is proportional to the square of the time $t$,we have $s \propto t^{2}$.This can be written as $s = k t^{2}$,where $k$

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Constant but not Zero

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(B) Given that the distance $s$ is proportional to the square of the time $t$,we have $s \propto t^{2}$.This can be written as $s = k t^{2}$,where $k$ is a constant.The velocity $v$ is the first derivative of distance with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(k t^{2}) = 2kt$.The acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(2kt) = 2k$.Since $k$ is a constant,$2k$ is also a constant and is not equal to zero.Therefore,the acceleration of the body is constant but not zero.

$A$ moving body is covering distances which are proportional to the square of the time. Then the acceleration of the body is

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