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

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Uniform acceleration

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(A) Given that the distance $s$ is proportional to the square of time $t$,we have $s \propto t^2$.This can be written as $s = kt^2$,where $k$ is a constant.The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(kt^2) = 2kt$.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,the acceleration $a = 2k$ is also a constant. Therefore,the particle travels with uniform acceleration.

The distance travelled by a particle is proportional to the square of time. Then,the particle travels with:

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