Tangential acceleration of a particle moving | vedclass

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Question

using equation for a straight line $(y=m x+c)$ where, $m=	an 60^{\circ}=\sqrt{3}(\mathrm{slop}) C=0$ from graph

relation b/w $a_{T}$ and $t$ is ob

Vedclass · Accepted Answer

$2^{2/3} \,\,sec$

Vedclass · Answer

using equation for a straight line $(y=m x+c)$ where, $m=	an 60^{\circ}=\sqrt{3}(\mathrm{slop}) C=0$ from graph

relation b/w $a_{T}$ and $t$ is obtained as

$a_{T}=	an 60^{\circ} t+0$

this tangential acceleration increases velocity $(v)$ of the particless as.

$a_{T}=\frac{d v}{d t} \Rightarrow \frac{d v}{d t}=\sqrt{3} t$

$\Rightarrow \int d v=\int \sqrt{3}+d t \Rightarrow v=\frac{\sqrt{3} t^{2}}{2}$

so, $a_{T}$ a particular time, centripetal accin $\left(a_{c}ight)$ is given by $a_{c}=\frac{v^{2}}{r}=\frac{3}{4} t^{4}(r=1) \vec{a}_{T}$

let a time $\vec{a}$ masses angle $30^{\circ}$ with $\vec{a}_{c} .$ so it makes angles $60^{\circ}$ with $\overrightarrow{a_{T}}(\overrightarrow{a_{c}} K \overrightarrow{a_{T}} are\,pendicular)$

$\Rightarrow \overrightarrow{a_{T}}=\sqrt{3 t^{2}+\frac{9}{16}+8 \frac{1}{2}}$ that given

$\Rightarrow 3 t^{2}=\frac{1}{11}\left(3 t^{2}+\frac{9}{16}+8ight)$

$\Rightarrow t=0$ or $t=2^{2 / 3}$

Tangential acceleration of a particle moving in a circle of radius $1$ $m$ varies with time $t$ as (initial velocity of particle is zero). Time after which total acceleration of particle makes and angle of $30^o$ with radial acceleration is

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