A moving body is covering a distance directly | vedclass

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(D) Given that the distance $s$ is directly proportional to the square of time $t$,we can write: $s \propto t^2$ or $s = kt^2$,where $k$ is a constant

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constant

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(D) Given that the distance $s$ is directly proportional to the square of time $t$,we can write: $s \propto t^2$ or $s = kt^2$,where $k$ is a constant.To find the velocity $v$,we differentiate the distance with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(kt^2) = 2kt$.To find the acceleration $a$,we differentiate the velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(2kt) = 2k$.Since $k$ is a constant,$2k$ is also a constant.Therefore,the acceleration of the body is constant.

$A$ moving body is covering a distance directly proportional to the square of time. The acceleration of the body is

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