The x and y coordinates of a particle moving in a plane are given by x(t)=a (p t) and y(t)=b (p t) where a, b( a) and p are positive constants of appropriate dimensions and t is time. Then, which of the following is not

Correct answer: D. Distance travelled by the particle in time interval between t=0 and t=(π)/(2 P) is a aligned & x=a (p t), y=b (p t) & (x^2)/(a^2)+(y^2)/(b^2)=1, i.e. equation of ellipse & Now, r=x i + j =a (p t) i +b (p t j & v=(d r)/(d t)=-p asin(p t) i +p b (p t) j & a=(d v)/(d t)=-p^2 a (p t) i -p^2 b (p t) j & at t=(π)/(2 p) & v=-p a i and a=-p^2 b j aligned v perpendicular to a Also, a=-p^2 r , i.e. directed towards a fixed point

JIPMER 2017

Exam: JIPMER
Year: 2017
Subject: Physics
Chapter: Motion in Two Dimensions

JIPMER 2017 · Physics · Motion in Two Dimensions

The x and y coordinates of a particle moving in a plane are given...

The

x

and

y

coordinates of a particle moving in a plane are given by

x(t)=a \cos (p t)

and

y(t)=b \sin (p t)

where

a, b( \lt a)

and

p

are positive constants of appropriate dimensions and

t

is time. Then, which of the following is not true?

Options

AThe path of the particle is an ellipse.
BVelocity and acceleration of the particle are perpendicular to each other at $t=\frac{\pi}{2 p}$
CAcceleration of the particle is always directed towards a fixed point.
DDistance travelled by the particle in time interval between $t=0$ and $t=\frac{\pi}{2 P}$ is a

Correct answer

D. Distance travelled by the particle in time interval between $t=0$ and $t=\frac{\pi}{2 P}$ is a

Step-by-step solution

\begin{aligned} & x=a \cos (p t), y=b \sin (p t) \\ & \therefore \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \text{ i.e. equation of ellipse } \\ & \text{ Now, } r=x \hat{\mathbf{i}}+\hat{\mathbf{j}}=a \cos (p t) \hat{\mathbf{i}}+b \sin (p t \hat{\mathbf{j}} \\ & v=\frac{d r}{d t}=-p \operatorname{asin}(p t) \hat{\mathbf{i}}+p b \cos (p t) \hat{\mathbf{j}} \\ & a=\frac{d v}{d t}=-p^2 a \cos (p t) \hat{\mathbf{i}}-p^2 b \sin (p t) \hat{\mathbf{j}} \\ & \text{ at } t=\frac{\pi}{2 p} \\ & v=-p a \hat{\mathbf{i}} \text{ and } \mathbf{a}=-p^2 b \hat{\mathbf{j}} \end{aligned}

v \text{ perpendicular to a } Also,

a=-p^2 r

, i.e. directed towards a fixed point$

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