The velocity of a particle moving in a curvil | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $t = 1 \ s$:

Similar Questions

Two bodies are thrown up at angles of $45^\circ$ and $60^\circ$,respectively,with the horizontal. If both bodies attain the same vertical height,then the ratio of the velocities with which these are thrown is:

$A$ particle moves with a constant speed of $5 \, ms^{-1}$ as shown in the figure. What is the change in velocity during half a revolution in $ms^{-1}$?

$A$ particle moves with constant angular velocity in a circle. During the motion,its:

$A$ body of mass $100 \, g$ is projected at an angle of $30^\circ$ with the horizontal with a velocity of $20 \, m \, s^{-1}$. What is the change in its momentum at the maximum height in $kg \, m \, s^{-1}$?

Two bodies are projected from points $(0,0)$ and $(\sqrt{3}-1,0)$ with velocities $10 \ ms^{-1}$ and $v \ ms^{-1}$ respectively,as shown in the figure. The time after which they collide in space is . . . . . . . (in $s$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module