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Chapter 2: Problem 308

A particle in \(\mathrm{xy}\) plane is governed by $\mathrm{x}=\mathrm{A} \cos \omega \mathrm{t}, \mathrm{y}=\mathrm{A}$ \((1-\sin \omega \mathrm{t}) . \mathrm{A}\) and \(\omega\) are constants. What is the speed of the particle. (A) A\omegat (B) \(\mathrm{A} \omega^{2} \mathrm{t}\) (C) \(\mathrm{A} \omega\) (D) \(\mathrm{A}^{2} \omega \sin (\omega \mathrm{t} / 2)\)

Short Answer

Expert verified
The short answer is: The speed of the particle is (C) \(A\omega\).

Step by step solution

01

Find the velocity in the x-direction

To find the velocity in the x-direction, we need to take the time derivative of the x(t) equation: \(v_x(t) = \frac{dx}{dt}\). Using the chain rule, we have: \(v_x(t) = \frac{d}{dt} [A \cos(\omega t)] = -A\omega\sin(\omega t)\)
02

Find the velocity in the y-direction

Similarly for the y-direction, we need to take the time derivative of the y(t) equation: \(v_y(t) =\frac{dy}{dt}\). Using the chain rule, we have: \(v_y(t) = \frac{d}{dt} [A(1 - \sin(\omega t))] = -A\omega\cos(\omega t)\)
03

Find the speed of the particle

Now that we have the velocity components, we can find the speed of the particle using the formula: \(\mathrm{speed} = \sqrt{v_x(t)^2 + v_y(t)^2}\) Substitute \(v_x(t) = -A\omega\sin(\omega t)\) and \(v_y(t) = -A\omega\cos(\omega t)\): \(\mathrm{speed} = \sqrt{(-A\omega\sin(\omega t))^2 + (-A\omega\cos(\omega t))^2}\)
04

Simplify the speed equation

Expanding and simplifying the speed equation, we get: \(\mathrm{speed} = \sqrt{(A^2\omega^2\sin^2(\omega t) + A^2\omega^2\cos^2(\omega t))}\) Factoring out \(A^2\omega^2\), we have: \(\mathrm{speed} = \sqrt{A^2\omega^2(\sin^2(\omega t) + \cos^2(\omega t))}\) Since \(\sin^2(\omega t) + \cos^2(\omega t) = 1\), the speed expression simplifies to: \(\mathrm{speed} = A\omega\) Therefore, the correct answer is (C) \(A\omega\).

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Most popular questions from this chapter

Rohit completes a semicircular path of radius \(\mathrm{R}\) in 10 seconds. Calculate average speed and average velocity in \(\mathrm{ms}^{-1}\) (A) \([(2 \pi R) / 10],(2 R / 10)\) (B) \((\pi R / 10),(\mathrm{R} / 10)\) (C) \((\pi R / 10),(2 R / 10)\) (D) \([(2 \pi R) / 10],(\mathrm{R} / 10)\)

Find a unit vector in direction of \(\hat{1}+2 \hat{\jmath}-3 \mathrm{k}\) (A) \((1 / \sqrt{7})(\hat{1}+2 \hat{\jmath}-3 \mathrm{k})\) (B) \(-(1 / 2)(\hat{1}+2 \hat{\jmath}-3 \mathrm{k})\) (C) \((1 / \sqrt{14})(\hat{1}+2 \hat{j}-3 \mathrm{k})\) (D) \((1 / \sqrt{5})(\hat{1}+2 \hat{\jmath}-3 \mathrm{k})\)

The relation between velocity and position of a particle is \(\mathrm{V}=\mathrm{Ax}+\mathrm{B}\) where \(\mathrm{A}\) and \(\mathrm{B}\) are constants. Acceleration of the particle is \(10 \mathrm{~ms}^{-2}\) when its velocity is \(\mathrm{V}\), How much is the acceleration when its velocity is $2 \mathrm{~V}$ (A) \(20 \mathrm{~ms}^{-2}\) (B) \(10 \mathrm{~ms}^{-1}\) (C) \(5 \mathrm{~ms}^{-2}\) (D) 0

\(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) are nonzero vectors. Which from the followings is true ? (A) \(\left|\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right|^{2}-\left|\mathrm{A}^{\rightarrow}-\mathrm{B}^{\rightarrow}\right|^{2}=2\left(\mathrm{~A}^{2}+\mathrm{B}^{2}\right)\) (B) \(\left|\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right|^{2}-\left|\mathrm{A}^{\rightarrow}-\mathrm{B}^{\rightarrow}\right|^{2}=2\left(\mathrm{~A}^{2}+\mathrm{B}^{2}\right)\) (C) \(\left|\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right|^{2}-\left|\mathrm{A}^{\rightarrow}-\mathrm{B}^{\rightarrow}\right|^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}\) (D) \(\left|\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right|^{2}-\left|\mathrm{A}^{\rightarrow}-\mathrm{B}^{\rightarrow}\right|^{2}=\mathrm{A}^{2}-\mathrm{B}^{2}\)

Motion of a particle is described by \(\mathrm{x}=(\mathrm{t}-2)^{2}\) Find its velocity when it passes through origin. (A) \(0 \mathrm{~m} / \mathrm{s}\) (B) \(2 \mathrm{~ms}^{-1}\) (C) \(4 \mathrm{~ms}^{-1}\) (D) \(8 \mathrm{~ms}^{-1}\)
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