Kinematics Question 494
Question: The coordinates of a particle moving in a plane are given by $ =-\ 8\ m/s^{2}. $ and $ y(t)=b\sin (pt) $ where $ a,b(<a) $ and $ p $ are positive constants of appropriate dimensions. Then[IIT-JEE 1999]
Options:
A) The path of the particle is an ellipse
B) The velocity and acceleration of the particle are normal to each other at $ t=\pi /(2p) $
C) The acceleration of the particle is always directed towards a focus
D) The distance travelled by the particle in time interval $ t=0 $ to $ t=\pi /(2p) $ it’s $ a $
Show Answer
Answer:
Correct Answer: A
Solution:
$ x=a\cos (pt) $ and $ y=b\sin (pt) $ (given) $ \therefore $ $ \cos pt=\frac{x}{a} $ and $ \sin pt=\frac{y}{b} $
By squaring and adding $ {{\cos }^{2}}(pt)+{{\sin }^{2}}(pt)=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $
Hence path of the particle is ellipse.
Now differentiating x and y w.r.t. time $ v _{x}=\frac{dx}{dt}=\frac{d}{dt}(a\cos (pt))=-ap\sin (pt) $
$ v _{y}=\frac{dy}{dt}=\frac{d}{dt}(b\sin (pt))=bp\cos (pt) $
$ \therefore \ \ \vec{v}=v _{x}\hat{i}+v _{y}\hat{j}=-ap\sin (pt)\hat{i}+bp\cos (pt)\hat{j} $
Acceleration
$ \vec{a}=\frac{d\vec{v}}{dt}=\frac{d}{dt}[-ap\sin (pt)\hat{i}+bp\cos (pt)\hat{j}] $
$ \vec{a}=-ap^{2}\cos (pt)\ \hat{i}-bp^{2}\sin (pt)\hat{j} $ Velocity at $ t=\frac{\pi }{2p} $
$ \vec{v}=-ap\sin p( \frac{\pi }{2p} )\ \hat{i}+bp\cos p( \frac{\pi }{2p} )\hat{j} $
$ =-ap\ \hat{i} $ Acceleration at $ t=\frac{\pi }{2p} $
$ \vec{a}=ap^{2}\cos p( \frac{\pi }{2p} )\ \hat{i}-bp^{2}\sin p( \frac{\pi }{2p} )\hat{j} $
$ =-bp^{2}\hat{j} $ As $ \vec{v}\ .\ \vec{a}=0 $ Hence velocity and acceleration are perpendicular to each other at $ t=\frac{\pi }{2p} $ .
