A particle moves in the xy-plane with velocit | vedclass

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

Question

(A) Given,velocity of particle is $\vec{v} = x \hat{i} + yt \hat{j}$.The magnitude of velocity is $v = \sqrt{x^2 + y^2 t^2}$.The tangential accelerati

Vedclass · Accepted Answer

$\frac{\sqrt{3} y}{2}, \frac{y}{2}$

Vedclass · Answer

(A) Given,velocity of particle is $\vec{v} = x \hat{i} + yt \hat{j}$.The magnitude of velocity is $v = \sqrt{x^2 + y^2 t^2}$.The tangential acceleration $a_t$ is the rate of change of speed:$a_t = \frac{dv}{dt} = \frac{1}{2\sqrt{x^2 + y^2 t^2}} \cdot (2y^2 t) = \frac{y^2 t}{\sqrt{x^2 + y^2 t^2}}$.Substituting $t = \frac{x \sqrt{3}}{y}$:$a_t = \frac{y^2 (x \sqrt{3} / y)}{\sqrt{x^2 + y^2 (3x^2 / y^2)}} = \frac{xy \sqrt{3}}{\sqrt{x^2 + 3x^2}} = \frac{xy \sqrt{3}}{2x} = \frac{\sqrt{3} y}{2}$.The total acceleration vector is $\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(x \hat{i} + yt \hat{j}) = y \hat{j}$.The magnitude of total acceleration is $a = |\vec{a}| = y$.The normal acceleration $a_n$ is given by $a_n = \sqrt{a^2 - a_t^2}$.$a_n = \sqrt{y^2 - (\frac{\sqrt{3} y}{2})^2} = \sqrt{y^2 - \frac{3y^2}{4}} = \sqrt{\frac{y^2}{4}} = \frac{y}{2}$.Thus,the tangential and normal accelerations are $\frac{\sqrt{3} y}{2}$ and $\frac{y}{2}$ respectively.

$A$ particle moves in the $xy$-plane with velocity $\vec{v} = x \hat{i} + yt \hat{j}$. At $t = \frac{x \sqrt{3}}{y}$,the tangential and normal accelerations are:

Similar Questions

In non-uniform circular motion,the ratio of tangential to radial acceleration is ($r=$ radius of the circle,$u=$ speed of the particle,$\alpha=$ angular acceleration).

$A$ body is moving on a circle of radius $80 \, m$ with a speed $20 \, m/s$ which is decreasing at the rate of $5 \, m/s^2$ at an instant. The angle made by its acceleration with its velocity is .......... (in $^{\circ}$)

$A$ car is moving at a speed of $40\,m/s$ on a circular track of radius $400\,m.$ This speed is increasing at the rate of $3\,m/s^2.$ The acceleration of the car is ........ $m/s^2$.

Explain the effects of the radial component and the tangential component of the linear acceleration of a particle in circular motion.

$A$ point object moves along an arc of a circle of radius $R$. Its velocity depends upon the distance covered $s$ as $v=K \sqrt{s}$,where $K$ is a constant. If $\theta$ is the angle between the total acceleration and tangential acceleration,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module