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(A) Given that the displacement $s$ is proportional to the cube of time $t$,we can write $s = kt^3$,where $k$ is a constant.To find the velocity $v$,w

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increasing with time

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(A) Given that the displacement $s$ is proportional to the cube of time $t$,we can write $s = kt^3$,where $k$ is a constant.To find the velocity $v$,we differentiate $s$ with respect to time $t$:$v = \frac{ds}{dt} = \frac{d}{dt}(kt^3) = 3kt^2$.To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:$a = \frac{dv}{dt} = \frac{d}{dt}(3kt^2) = 6kt$.Since $a = 6kt$,the acceleration $a$ is directly proportional to time $t$. Therefore,the magnitude of the acceleration increases linearly with time.

Time	Distance from her home $(km)$
$06:45 \, am$	$0$
$07:00 \, am$	$8$
$01:30 \, pm$	$8$
$01:45 \, pm$	$0$

The displacement of a body is proportional to the cube of the time lapsed. The magnitude of the acceleration is

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Time Distance from her home $(km)$

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$07:00 \, am$ $8$

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$01:45 \, pm$ $0$

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