The displacement of a particle is proportiona | vedclass

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Question

(d) $x \propto {t^3}$
$	herefore $$x = K{t^3}$
$ \Rightarrow v = \frac{{dx}}{{dt}} = 3\;K{t^2}$ and $ \Rightarrow a = \frac{{dv}}{{dt}} = 6\;Kt$
i

Vedclass · Accepted Answer

$a \propto t$

Vedclass · Answer

(d) $x \propto {t^3}$
$	herefore $$x = K{t^3}$
$ \Rightarrow v = \frac{{dx}}{{dt}} = 3\;K{t^2}$ and $ \Rightarrow a = \frac{{dv}}{{dt}} = 6\;Kt$
i.e. $a \propto t$

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

The displacement of a particle is proportional to the cube of time elapsed. How does the acceleration of the particle depends on time obtained

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