A particle moves along a straight line such t | vedclass

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Question

$s=t^{3}-6 t^{2}+3 t+4$
$\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=3 \mathrm{t}^{2}-12 \mathrm{t}+3$
$a=\frac{d v}{d t}=6 t-12$
$a=0$
$t=2 \sec$

Vedclass · Accepted Answer

$-9$

Vedclass · Answer

$s=t^{3}-6 t^{2}+3 t+4$
$\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=3 \mathrm{t}^{2}-12 \mathrm{t}+3$
$a=\frac{d v}{d t}=6 t-12$
$a=0$
$t=2 \sec$
Velocity at $t=2 \mathrm{sec}$
$\mathrm{v}=-9 \mathrm{m} / \mathrm{s}$

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

A particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^3 -6t^2 + 3t + 4)\, m$. ...... $m/s$ is the velocity of the particle when its acceleration is zero ................. $\mathrm{m/s}$

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