The velocity of a particle moving in a curvil | vedclass

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(D) Given velocity: $\vec{v} = 2t\hat{i} + t^2\hat{j}$.Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = 2\hat{i} + 2t\hat{j}$.At $t = 1 \ s$,$\vec{a} = 2

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given velocity: $\vec{v} = 2t\hat{i} + t^2\hat{j}$.Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = 2\hat{i} + 2t\hat{j}$.At $t = 1 \ s$,$\vec{a} = 2\hat{i} + 2\hat{j}$,so magnitude $a = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \ m/s^2$.Speed $v = |\vec{v}| = \sqrt{(2t)^2 + (t^2)^2} = \sqrt{4t^2 + t^4}$.At $t = 1 \ s$,$v = \sqrt{4+1} = \sqrt{5} \ m/s$.Tangential acceleration $a_t = \frac{dv}{dt} = \frac{1}{2\sqrt{4t^2 + t^4}} \cdot (8t + 4t^3)$.At $t = 1 \ s$,$a_t = \frac{8+4}{2\sqrt{5}} = \frac{12}{2\sqrt{5}} = \frac{6}{\sqrt{5}} \ m/s^2$.Radial (centripetal) acceleration $a_c = \sqrt{a^2 - a_t^2} = \sqrt{8 - \frac{36}{5}} = \sqrt{\frac{40-36}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} \ m/s^2$.Radius of curvature $R = \frac{v^2}{a_c} = \frac{(\sqrt{5})^2}{2/\sqrt{5}} = \frac{5 \cdot \sqrt{5}}{2} = \frac{5\sqrt{5}}{2} \ m$.Thus,option $D$ is correct.

The 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 seconds. At $t = 1 \ s$:

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