The velocity of a particle moving on the x- a | vedclass

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Question

$\mathrm{V}=\mathrm{x}^{2}+\mathrm{x} \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}+\frac{\mathrm{dx}}{\mat

Vedclass · Accepted Answer

$30$

Vedclass · Answer

$\mathrm{V}=\mathrm{x}^{2}+\mathrm{x} \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}+\frac{\mathrm{dx}}{\mathrm{dt}}$
$\Rightarrow \mathrm{a}=(2 \mathrm{x}+1) \mathrm{V}=(2 \mathrm{x}+1)\left(\mathrm{x}^{2}+\mathrm{x}ight)$
$	herefore$ When $\mathrm{x}=2 \mathrm{m}$
$a=(2 	imes 2+1)\left(2^{2}+2ight)=5 	imes 6=30 \mathrm{m} / \mathrm{s}^{2}$

Colum $I$	Colum $II$
$(A)$ Distance travelled in $3\,s$	$(p)$ $-20$ units
$(B)$ Displacement in $1\,s$	$(q)$ $15$ units
$(C)$ Initial acceleration	$(r)$ $25$ units
$(D)$ Velocity at $4\,s$	$(s)$ $-10$ units

The velocity of a particle moving on the $x-$ axis is given by $v = x^2 + x$ where $v$ is in $m/s$ and $x$ is in $m$ . Find its acceleration in $m/s^2$ when passing through the point $x = 2m$

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$(B)$ Displacement in $1\,s$ $(q)$ $15$ units

$(C)$ Initial acceleration $(r)$ $25$ units

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