The distance travelled by a body moving along | vedclass

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(B) Given that the distance (or displacement) $s$ is proportional to $t^3$,we can write:$s = k t^3$ (where $k$ is a constant).To find the velocity $v$

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(B) Given that the distance (or displacement) $s$ is proportional to $t^3$,we can write:$s = k t^3$ (where $k$ is a constant).To find the velocity $v$,we differentiate $s$ with respect to time $t$:$v = \frac{ds}{dt} = \frac{d}{dt}(k t^3) = 3k t^2$.To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:$a = \frac{dv}{dt} = \frac{d}{dt}(3k t^2) = 6k t$.Since $a = 6k t$,it implies that $a \propto t$.This represents a linear relationship passing through the origin,which corresponds to a straight line graph starting from the origin.Looking at the provided options,the graph that shows $a$ increasing linearly with $t$ is Graph $B$.

$t (s)$	$0$	$1$	$2$	$3$
$x (m)$	$-2$	$0$	$6$	$16$

The distance travelled by a body moving along a line in time $t$ is proportional to $t^3$. The acceleration-time $(a, t)$ graph for the motion of the body will be

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