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Chapter 2: Problem 249

An object moves in \(\mathrm{x}-\mathrm{y}\) plane. Equations for displacement in \(\mathrm{x}\) and \(\mathrm{y}\) direction are $\mathrm{x}=3 \sin 2 \mathrm{t}\( and \)\mathrm{y}=3 \cos 2 \mathrm{t}$ Speed of the particle is (A) zero (B) constant and nonzero (C) increasing with time t (D) decreasing with time t

Short Answer

Expert verified
The speed of the particle is found by differentiating the displacement equations with respect to time and then finding the magnitude of the resulting velocity vector. After doing this, the speed is found to be a constant value of 6, which means the answer is (B) constant and nonzero.

Step by step solution

01

Differentiate the displacement equations

Differentiate the displacement equations with respect to time, to find the velocities in the x and y direction: \(v_x = \frac{dx}{dt} = 3 \cdot 2 \cos{(2t)}\) \(v_y = \frac{dy}{dt} = -3 \cdot 2 \sin{(2t)}\)
02

Find the speed

Now that we have the velocities in the x and y direction, we can find the speed, which is the magnitude of the velocity vector. So the speed \(v\) is: \(v = \sqrt{v_x^2 + v_y^2}\)
03

Plug in the calculated velocities

Plugging in the calculated velocities, \(v_x\) and \(v_y\), into the speed equation from step 2: \(v = \sqrt{(3 \cdot 2 \cos{(2t)})^2 + (-3 \cdot 2 \sin{(2t)})^2}\)
04

Simplify the equation

Simplify the speed equation: \(v = \sqrt{36 \cos^2{(2t)} + 36 \sin^2{(2t)}}\) Factor out 36: \(v = \sqrt{36(\cos^2{(2t)} + \sin^2{(2t)})}\) Now, since the Pythagorean identity states that \(\sin^2{x} + \cos^2{x} = 1\) for any x, the expression inside the parentheses becomes 1: \(v = \sqrt{36}\) So, \(v = 6\), which is a constant value. Now that we have found the speed of the object, we can conclude that the answer is: (B) constant and nonzero.

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Most popular questions from this chapter

\(\mathrm{P}^{\rightarrow}=\mathrm{Q}^{\rightarrow}\) is true, if \(\ldots\) (A) their magnitudes are equal (B) they are in same direction (C) their magnitudes are equal and they are in same direction (D) their magnitudes are not equal and they are not in same direction

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) \(8 \mathrm{~ms}^{-1}\)

\(\mathrm{A}^{\rightarrow}\) 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) \(\left|\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right|^{2}-\left|\mathrm{A}^{\rightarrow}-\mathrm{B}^{\rightarrow}\right|^{2}=\mathrm{A}^{2}-\mathrm{B}^{2}\)

\(\mathrm{P}^{-}\) and \(\mathrm{Q}^{-}\) are equal vectors what from the followings is true. (A) \(\mathrm{P}^{-}\) and \(\mathrm{Q}^{-}\) are anti parallel (B) \(\mathrm{P}^{-}\) and \(\mathrm{Q}^{-}\) are parallel (C) \(\mathrm{P}^{-}\) and \(\mathrm{Q}^{-}\) may be perpendicular (D) \(\mathrm{P}^{-}\) and \(\mathrm{Q}^{\rightarrow}\) may be free vectors

A particle moves on a plane along the path \(\mathrm{y}=\mathrm{Ax}^{3}+\mathrm{B}\) in such a way that $(\mathrm{dx} / \mathrm{dt})=\mathrm{c} . \mathrm{c}, \mathrm{A}, \mathrm{B}$ are constant. Calculate the acceleration of the particle. (A) \(3 \mathrm{Ax} \mathrm{cj} \mathrm{ms}^{-2}\) (B) \(6 \mathrm{Axc}^{2} \hat{\mathrm{j}} \mathrm{ms}^{-2}\) (C) \(3 \mathrm{Axc}^{2} \hat{\mathrm{j}} \mathrm{ms}^{-2}\) (D) \(\left[c \hat{i}+3 A x c^{2} \hat{j}\right] m s^{-2}\)
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