SATHEE: Kinematics Question 494

Kinematics Question 494

Question: The coordinates of a particle moving in a plane are given by $= - 8 m / s^{2} .$ and $y (t) = b \sin (p t)$ 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 $t = π / (2 p)$

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 $t = π / (2 p)$ it’s $a$

Show Answer

Answer:

Correct Answer: A

Solution:

$x = a \cos (p t)$ and $y = b \sin (p t)$ (given) $∴$ $\cos p t = \frac{x}{a}$ and $\sin p t = \frac{y}{b}$

By squaring and adding $\cos^{2} (p t) + \sin^{2} (p t) = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Hence path of the particle is ellipse.

Now differentiating x and y w.r.t. time $v_{x} = \frac{d x}{d t} = \frac{d}{d t} (a \cos (p t)) = - a p \sin (p t)$

$v_{y} = \frac{d y}{d t} = \frac{d}{d t} (b \sin (p t)) = b p \cos (p t)$

$∴ \vec{v} = v_{x} \hat{i} + v_{y} \hat{j} = - a p \sin (p t) \hat{i} + b p \cos (p t) \hat{j}$

Acceleration

$\vec{a} = \frac{d \vec{v}}{d t} = \frac{d}{d t} [- a p \sin (p t) \hat{i} + b p \cos (p t) \hat{j}]$

$\vec{a} = - a p^{2} \cos (p t) \hat{i} - b p^{2} \sin (p t) \hat{j}$ Velocity at $t = \frac{π}{2 p}$

$\vec{v} = - a p \sin p (\frac{π}{2 p}) \hat{i} + b p \cos p (\frac{π}{2 p}) \hat{j}$

$= - a p \hat{i}$ Acceleration at $t = \frac{π}{2 p}$

$\vec{a} = a p^{2} \cos p (\frac{π}{2 p}) \hat{i} - b p^{2} \sin p (\frac{π}{2 p}) \hat{j}$

$= - b p^{2} \hat{j}$ As $\vec{v} . \vec{a} = 0$ Hence velocity and acceleration are perpendicular to each other at $t = \frac{π}{2 p}$ .

Kinematics Question 494

Question: The coordinates of a particle moving in a plane are given by $= - 8 m / s^{2} .$ and $y (t) = b \sin (p t)$ where $a, b (< a)$ and $p$ are positive constants of appropriate dimensions. Then[IIT-JEE 1999]

Options:

Answer:

Solution:

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Mentorship

NCERT Exercise Video Solution

Kinematics Question 494

Question: The coordinates of a particle moving in a plane are given by =− 8 m/s2. and y(t)=bsin⁡(pt) where a,b(<a) and p are positive constants of appropriate dimensions. Then[IIT-JEE 1999]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: The coordinates of a particle moving in a plane are given by $= - 8 m / s^{2} .$ and $y (t) = b \sin (p t)$ where $a, b (< a)$ and $p$ are positive constants of appropriate dimensions. Then[IIT-JEE 1999]