"Displacement is directly proportional to the square of the time elapsed". Is the acceleration constant or variable? Justify your answer.

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Solution

Displacement $x = k t^{2}$
where $k$ is a constant.
Velocity $v = \frac{d x}{d t} = 2 k t$
Acceleration $a = \frac{d v}{d t} = 2 k = c o n s t a n t$

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"Displacement is directly proportional to the square of the time elapsed". Is the acceleration constant or variable? Justify your answer.

Displacement x=kt2where k is a constant.Velocity v=dxdt=2ktAcceleration a=dvdt=2k=constant

Displacement $x = k t^{2}$
where $k$ is a constant.
Velocity $v = \frac{d x}{d t} = 2 k t$
Acceleration $a = \frac{d v}{d t} = 2 k = c o n s t a n t$