Problem 19 - Calculus Early Transcendentals

Question

The function \(s(t)\) describes the position of a particle moving along a coordinate line, where \(s\) is in feet and \(t\) is in seconds. (a) Find the velocity and acceleration functions. (b) Find the position, velocity, specd, and acceleration at time \(t=1 .\) (c) At what times is the particle stopped? (d) When is the particle speeding up? Slowing down? (e) Find the total distance traveled by the particle from time \(t=0\) to time \(t=5\). $$ s(t)=9-9 \cos (\pi t / 3), \quad 0 \leq t \leq 5 $$

Step-by-Step Solution

Answer

(a) Velocity: \( v(t) = 3\pi \sin(\frac{\pi t}{3}) \); Acceleration: \( a(t) = \pi^2 \cos(\frac{\pi t}{3}) \). (b) \( s(1) = 4.5 \) ft; \( v(1) \approx 8.168 \) ft/s; \( a(1) \approx 4.935 \) ft/s². (c) Particle stops at \( t = 0, 3 \). (d) Speeding up: \( (0, 1.5) \), \( (3, 4.5) \); Slowing down: \( (1.5, 3) \), \( (4.5, 5) \). (e) Total distance: 18 ft (full cycles or integral confirmation).

Step 1: Calculating the Velocity Function

The velocity function, denoted as \( v(t) \), is the derivative of the position function \( s(t) \) with respect to time \( t \). We'll use the chain rule to differentiate: \[ v(t) = \frac{d}{dt}(9 - 9 \cos(\pi t / 3)) = 9 \times \frac{d}{dt}(-\cos(\pi t / 3)) = 9 \times (\sin(\pi t / 3)) \times \frac{\pi}{3} = 3\pi \sin(\pi t / 3) \]

Step 2: Calculating the Acceleration Function

The acceleration function, denoted as \( a(t) \), is the derivative of the velocity function \( v(t) \). We'll differentiate \( v(t) = 3\pi \sin(\pi t / 3) \):\[ a(t) = \frac{d}{dt}(3\pi \sin(\pi t / 3)) = 3\pi \times \cos(\pi t / 3) \times \frac{\pi}{3} = \pi^2 \cos(\pi t / 3) \]

Step 3: Finding Position, Velocity and Acceleration at \( t = 1 \)

First, find \( s(1) \):\[ s(1) = 9 - 9\cos(\pi \times 1 / 3) = 9 - 9 \times \frac{1}{2} = 4.5 \text{ feet} \]Next, find \( v(1) \):\[ v(1) = 3\pi \sin(\pi \times 1 / 3) = 3\pi \times \frac{\sqrt{3}}{2} = \frac{3\pi\sqrt{3}}{2} \approx 8.168 \text{ ft/s} \]Finally, find \( a(1) \):\[ a(1) = \pi^2 \cos(\pi \times 1 / 3) = \pi^2 \times \frac{1}{2} = \frac{\pi^2}{2} \approx 4.935 \text{ ft/s}^2 \]

Step 4: Determining When the Particle is Stopped

A particle is stopped when its velocity is zero. Set the velocity function equal to zero:\[ 3\pi \sin(\pi t / 3) = 0 \]Dividing through by \(3\pi\), we get:\[ \sin(\pi t / 3) = 0 \]The general solution to this is \( \pi t / 3 = n\pi \), where \( n \) is an integer. Solving for \( t \), we have:\[ t = 3n \]For \( 0 \leq t \leq 5 \), \( t = 0, 3 \). Thus, the particle stops at \( t = 0 \) and \( t = 3 \).

Step 5: Assessing When the Particle is Speeding Up or Slowing Down

A particle is speeding up when the velocity and acceleration have the same sign, and slowing down when they have opposite signs. We examine the sign of \( v(t) \) and \( a(t) \) over \( 0 \leq t \leq 5 \):- For \( t \) in interval \( (0, 1.5) \), \( \sin(\pi t / 3) > 0 \) and \( \cos(\pi t / 3) > 0 \): Speeding up.- For \( t \) in interval \( (1.5, 3) \), \( \sin(\pi t / 3) > 0 \) and \( \cos(\pi t / 3) < 0 \): Slowing down.- For \( t \) in interval \( (3, 4.5) \), \( \sin(\pi t / 3) < 0 \) and \( \cos(\pi t / 3) < 0 \): Speeding up.- For \( t \) in interval \( (4.5, 5) \), \( \sin(\pi t / 3) < 0 \) and \( \cos(\pi t / 3) > 0 \): Slowing down.

Step 6: Calculating Total Distance Traveled from \( t = 0 \) to \( t = 5 \)

The total distance traveled is the integral of the absolute value of velocity over the interval. This requires evaluating \( |v(t)| \) through the sub-intervals where parabolic direction changes.Evaluate separately over intervals defined by zero crossings found in Step 4 and reassess speeding up or slowing down zones:1. \( 0 \to 3 \): Location \( s(3) = 18 \)2. \( 3 \to 5 \): Move from the result at \( t=3 \) to \( s(5)=9 \)Total change = Through direct calculation from known trigonometric properties, or graphical validation: 18 feet from 3 to stop at 5, requiring 9 - 18 feet.

Key Concepts

Differentiation

In calculus, differentiation is a core concept that deals with how we calculate the rate at which things change. When a function represents a quantity over time, like the position of a moving particle, differentiation allows us to find its velocity and acceleration by examining its changes over time. For example, if you differentiate the position function, you get the velocity function, which tells us how fast the position is changing. Similarly, differentiating the velocity function gives us the acceleration function. Differentiation makes use of rules such as the product, quotient, and chain rules. In our context, we've used the chain rule to differentiate the function representing position:

• Velocity: Differentiate position to get a function that describes how fast it changes.
• Acceleration: Differentiate velocity to get a function that describes how fast the speed is changing.

Velocity and Acceleration

Understanding velocity and acceleration is crucial when studying motion along a coordinate line. Velocity, denoted as \( v(t) \), reflects how the position of a particle changes with time. It can be positive, negative, or zero, indicating forward motion, backward motion, or a complete stop, respectively. Meanwhile, acceleration \( a(t) \) describes how the velocity changes over time. Both are derived from the position function \( s(t) \):

• Velocity: \( v(t) = 3\pi \sin(\pi t / 3) \).
• Acceleration: \( a(t) = \pi^2 \cos(\pi t / 3) \).

Velocity gives insight into the directional speed of a particle. Acceleration helps understand how quickly that speed is changing. For instance, when both velocity and acceleration have the same sign, the particle is speeding up. If the signs are opposite, the particle is slowing down. When the velocity reaches zero, it indicates the particle is momentarily stopped.

Trigonometric Functions

Trigonometric functions like sine and cosine play a vital role in defining periodic motions, which are common in particle movements. Here, the position of the particle is given by the function: \[ s(t)=9-9 \cos(\pi t / 3) \]. These functions oscillate between -1 and 1, making them perfect for modeling scenarios where values periodically increase and decrease, such as waves and circular motion. The derivatives, which are other trigonometric functions, provide us the velocity and acceleration:

• The derivative of cosine is negative sine, so when finding velocity, you see \( \sin(\pi t / 3) \).
• The derivative of sine is cosine, leading to \( \cos(\pi t / 3) \) for acceleration.

Understanding these functions helps us describe and anticipate patterns in periodic motions effectively.

Particle Motion

Particle motion refers to the changing position of a particle over time, which can be determined by functions of time, like our harmonic function for position. Such functions might represent a simple oscillating motion or a more complex path. Thinking about where and when particles stop gives insight into their motion:

• Stopped particles have zero velocity. In this scenario, the particle stops at specific time intervals \( t = 0 \) and \( t = 3 \).
• The motion involves speeding up and slowing down depending on the sign of velocity and acceleration; this relates to when their product is positive or negative.

Calculating total distance traveled involves integrating the velocity over a time period and accounting for direction changes, ensuring distance traveled is positive regardless of direction changes. This involves understanding the trigonometric aspect of motion as particles might pass through the same point several times in different directions within a given time.