Chapter 3: Problem 64
Assume that a particle's position on the \(x\) -axis is given by $$ x=3 \cos t+4 \sin t $$ where \(x\) is measured in feet and \(t\) is measured in seconds. a. Find the particle's position when \(t=0, t=\pi / 2,\) and \(t=\pi .\) b. Find the particle's velocity when \(t=0, t=\pi / 2,\) and \(t=\pi\)
Short Answer
Step by step solution
Find Position at t=0
Find Position at t=Ï€/2
Find Position at t=Ï€
Differentiate Position to Find Velocity Function
Find Velocity at t=0
Find Velocity at t=Ï€/2
Find Velocity at t=Ï€
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Position Function
- At \( t=0 \), you substitute zero into the position function: \[ x = 3 \cos(0) + 4 \sin(0) = 3 \times 1 + 4 \times 0 = 3 \text{ feet}. \]
- At \( t=\pi/2 \), substitute \( \pi/2 \): \[ x = 3 \cos(\pi/2) + 4 \sin(\pi/2) = 3 \times 0 + 4 \times 1 = 4 \text{ feet}. \]
- At \( t=\pi \), use \( \pi \) in the equation: \[ x = 3 \cos(\pi) + 4 \sin(\pi) = 3 \times (-1) + 4 \times 0 = -3 \text{ feet}. \]
Velocity Function
- At \( t=0 \), \[ v(0) = -3 \sin(0) + 4 \cos(0) = 0 + 4 = 4 \text{ feet per second}. \]
- At \( t=\pi/2 \), \[ v(\pi/2) = -3 \sin(\pi/2) + 4 \cos(\pi/2) = -3 + 0 = -3 \text{ feet per second}. \]
- At \( t=\pi \), \[ v(\pi) = -3 \sin(\pi) + 4 \cos(\pi) = 0 - 4 = -4 \text{ feet per second}. \]
Trigonometric Derivatives
- The derivative of \( \cos t \) is \( -\sin t \).
- The derivative of \( \sin t \) is \( \cos t \).
Simple Harmonic Motion
- The position function of an SHM can be noted as \( x(t) = A \cos(\omega t + \phi) \), where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase shift.
- Our position function \( x(t) = 3 \cos t + 4 \sin t \) is a variation of standard SHM, indicating the particle's oscillatory motion along the x-axis without additional angular frequency or phase shift.