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

$A$ particle moves on a circle of radius $0.1 \ m$ with a velocity $v = 1.0t$. The total acceleration at $t = 5 \ s$ will be ........ $m/s^2$.

$A$ 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 ........ $m/s^2$.

$A$ body of mass $2 \,kg$ moves in a horizontal circular path of radius $5 \,m$. At an instant, its speed is $2 \sqrt{5} \,m/s$ and is increasing at the rate of $3 \,m/s^2$. The magnitude of the net force acting on the body at that instant is: (in $\,N$)

$A$ particle moves in a circular path of radius $R$ with an angular velocity $\omega = a - bt$,where $a$ and $b$ are positive constants and $t$ is time. The magnitude of the acceleration of the particle after time $t = \frac{2a}{b}$ is:

$A$ stone is tied to one end of a string $50\, cm$ long and is whirled in a horizontal circle with a constant speed. If the stone makes $10$ revolutions in $20\, s$,what is the magnitude of the acceleration of the stone in $cm/s^2$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module