Velocity of a particle moving in a curvilinea | vedclass

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Question

$\overrightarrow{\mathrm{V}}=2 \mathrm{t} \hat{\mathrm{i}}+\mathrm{t}^{2} \hat{\mathrm{j}}$

$\vec{a}=2 \hat{i}+2 t \hat{j}$

$|\vec{V}|=V=\sqrt{4

Vedclass · Accepted Answer

radius of curvature to the path is $\frac{5\sqrt 5}{{2 }} \ m$

Vedclass · Answer

$\overrightarrow{\mathrm{V}}=2 \mathrm{t} \hat{\mathrm{i}}+\mathrm{t}^{2} \hat{\mathrm{j}}$

$\vec{a}=2 \hat{i}+2 t \hat{j}$

$|\vec{V}|=V=\sqrt{4 t^{2}+t^{4}}$

$a_{t}=\left(\frac{d V}{d t}ight)=\frac{1}{2 \sqrt{4 t^{2}+t^{4}}} 	imes\left(8 t+4 t^{3}ight)$

at $t=1 \quad a_{t}=\left(\frac{6}{\sqrt{5}}ight) \mathrm{m} / \mathrm{s}^{2}$

$a=\sqrt{2^{2}+(2 t)^{2}}=\sqrt{8} \mathrm{m} / \mathrm{s}^{2}$

$r=\sqrt{5}$

$a_{c}=\sqrt{a^{2}-a_{t}^{2}}=\sqrt{8-\frac{36}{5}}=\left(\frac{2}{\sqrt{5}}ight)$

$R=\left(\frac{V^{2}}{a_{c}}ight)=\frac{5}{\left(\frac{2}{\sqrt{5}}ight)}=\frac{5 \sqrt{5}}{2}$

$R=\left(\frac{5 \sqrt{5}}{2}ight) m$

Velocity of a particle moving in a curvilinear path in a horizontal $X$ $Y$ plane varies with time as $\vec v = (2t\hat i + t^2 \hat j) \ \ m/s.$ Here, $t$ is in second. At $t = 1\ s$

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