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(D) Given that displacement $s$ is proportional to the square of time $t$,we can write $s = kt^2$,where $k$ is a constant.Velocity $v$ is the rate of

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uniform acceleration

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(D) Given that displacement $s$ is proportional to the square of time $t$,we can write $s = kt^2$,where $k$ is a constant.Velocity $v$ is the rate of change of displacement with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(kt^2) = 2kt$.Acceleration $a$ is the rate of change of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(2kt) = 2k$.Since $2k$ is a constant,the acceleration of the object is constant or uniform.Therefore,the object moves with uniform acceleration.

If the displacement of an object is proportional to the square of time,then the object moves with

Similar Questions

Explain the following types of motion with one example for each:
$(i)$ Acceleration is positive
$(ii)$ Acceleration is negative
$(iii)$ Acceleration is zero.

What can you conclude about the motion of a body depicted by the velocity-time graphs $(i)$,$(ii)$ and $(iii)$ given below?

What is the nature of the displacement-time graph of a body moving with constant velocity?

Identify what the graphs shown below indicate.

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