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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 $B$.

Column $I$	Column $II$
$(A)$ $\frac{dv}{dt}$	$(p)$ Acceleration
$(B)$ $\frac{d\|v\|}{dt}$	$(q)$ Magnitude of acceleration
$(C)$ $\frac{dr}{dt}$	$(r)$ Velocity
$(D)$ $\left\|\frac{dr}{dt}\right\|$	$(s)$ Magnitude of velocity

Column-$I$	Column-$II$
$(1)$ Acceleration is positive	$(a)$ Speed of the particle decreases
$(2)$ Acceleration is negative	$(b)$ Speed of the particle increases
	$(c)$ Speed of the particle keeps changing

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

Similar Questions

$A$ particle moves along the $x$-axis and its displacement $x$ varies with time $t$ according to the equation $x = c_0(t^2 - 2) + c(t - 2)^2$,where $c_0$ and $c$ are constants of appropriate dimensions. Which of the following statements is correct?

Match the following columns.
Column $I$ Column $II$
$(A)$ $\frac{dv}{dt}$ $(p)$ Acceleration
$(B)$ $\frac{d|v|}{dt}$ $(q)$ Magnitude of acceleration
$(C)$ $\frac{dr}{dt}$ $(r)$ Velocity
$(D)$ $\left|\frac{dr}{dt}\right|$ $(s)$ Magnitude of velocity

Match the items in Column-$I$ with the items in Column-$II$ correctly.

Column-$I$ Column-$II$

$(1)$ Acceleration is positive $(a)$ Speed of the particle decreases

$(2)$ Acceleration is negative $(b)$ Speed of the particle increases

$(c)$ Speed of the particle keeps changing

The velocity $(v)-$ time $(t)$ plot of the motion of a body is shown below:
The acceleration $(a)-$ time $(t)$ graph that best suits this motion is:

If the displacement ($s$ in metre) of a moving particle in terms of time ($t$ in second) is $s = t^3 - 6t^2 + 18t + 9$,then the minimum velocity attained by the particle is (in $m \ s^{-1}$)

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