Problem 6 - Thomas Calculus

Question

Exercises \(1-6\) give the positions \(s=f(t)\) of a body moving on a coordinate line, with \(s\) in meters and \(t\) in seconds. a. Find the body's displacement and average velocity for the given time interval. b. Find the body's speed and acceleration at the endpoints of the interval. c. When, if ever, during the interval does the body change direction? $$ s=\frac{25}{t+5}, \quad-4 \leq t \leq 0 $$

Step-by-Step Solution

Answer

The displacement is -20 meters, average velocity is -5 m/s. Speed at endpoints is 1 m/s & 25 m/s, acceleration is 0.4 m/s² & 50 m/s². No direction change occurs.

Step 1: Calculate Displacement

Find the displacement of the body by evaluating the position function at the endpoints of the given time interval. The displacement is calculated as the difference between the position at the end of the interval and the position at the start.Given:\[ s(t) = \frac{25}{t+5} \]Evaluate at the endpoints:\[ s(0) = \frac{25}{0+5} = 5 \]\[ s(-4) = \frac{25}{-4+5} = 25 \]The displacement is:\[ \Delta s = s(0) - s(-4) = 5 - 25 = -20 \text{ meters} \]

Step 2: Calculate Average Velocity

Average velocity is the displacement divided by the time interval.\[ \text{Average velocity} = \frac{\Delta s}{\Delta t} = \frac{-20}{0 - (-4)} = \frac{-20}{4} = -5 \text{ m/s} \]

Step 3: Calculate Speed and Acceleration at Endpoints

To find speed at the endpoints, we determine the magnitude of the derivative \( v(t) = s'(t) \), and for acceleration, we need the second derivative \( a(t) = s''(t) \).First, find \( s'(t) \):\[ s'(t) = \frac{d}{dt} \left( \frac{25}{t+5} \right) = -\frac{25}{(t+5)^2} \]Evaluate at \( t = 0 \):\[ v(0) = -\frac{25}{5^2} = -1 \text{ m/s} \] (Speed is 1 m/s)\( s''(t) \):\[ s''(t) = \frac{d}{dt} \left( -\frac{25}{(t+5)^2} \right) = \frac{50}{(t+5)^3} \]Evaluate at \( t = 0 \):\[ a(0) = \frac{50}{5^3} = 0.4 \text{ m/s}^2 \]Evaluate \( s'(t) \) at \( t = -4 \):\[ v(-4) = -\frac{25}{1^2} = -25 \text{ m/s} \] (Speed is 25 m/s)Evaluate \( s''(t) \) at \( t = -4 \):\[ a(-4) = \frac{50}{1^3} = 50 \text{ m/s}^2 \]

Step 4: Determine Direction Change

A direction change happens where the velocity function changes sign. Since \( v(t) = -\frac{25}{(t+5)^2} \) and \( (t+5)^2 > 0 \) for all \( t \), the sign of \( v(t) \) does not change within \([-4, 0]\). Therefore, there is no direction change in this interval.

Key Concepts

Displacement

Displacement in calculus refers to the change in position of an object. It is important to remember that displacement is not the same as distance. While distance considers how much ground is covered, displacement is more concerned with the initial and final positions, taking into account direction.
To find the displacement over a time interval, subtract the initial position from the final position of the object. In our given problem, the function defining position is \( s(t) = \frac{25}{t+5} \).
Evaluate at the endpoints of the time interval \(-4 \leq t \leq 0\):

• Initial position \( s(-4) = \frac{25}{1} = 25 \text{ meters}\)
• Final position \( s(0) = \frac{25}{5} = 5 \text{ meters}\)

So, the displacement \( \Delta s \) is \( 5 - 25 = -20 \text{ meters}\). The negative sign indicates a movement in the opposite direction in the context of our coordinate system.

Average velocity

Average velocity gives an idea of how fast an object is moving over a specific time interval and in which direction. Unlike average speed, which only considers magnitude, average velocity includes direction because it uses displacement.
To calculate average velocity, divide the total displacement by the total time elapsed. In the exercise:

• Total displacement \( \Delta s = -20 \text{ meters}\)
• Total time \( \Delta t = 0 - (-4) = 4 \text{ seconds}\)

Average velocity is given by:
\[\text{Average velocity} = \frac{\Delta s}{\Delta t} = \frac{-20}{4} = -5 \text{ m/s}\]
The negative sign in velocity indicates movement in the opposite direction along the coordinate line within the given time frame.

Acceleration

Acceleration measures how quickly velocity changes over time. In this problem, acceleration is determined by finding the second derivative of the position function \( s(t) \). This second derivative, denoted as \( s''(t) \), gives us the acceleration function.
First, to find velocity \( v(t) \), take the first derivative:

• \( s'(t) = -\frac{25}{(t+5)^2} \)

Next, find acceleration by differentiating the velocity function:

• \( s''(t) = \frac{50}{(t+5)^3} \)

Now, evaluate the acceleration at the endpoints:
At \( t = 0 \), \( a(0) = \frac{50}{5^3} = 0.4 \text{ m/s}^2 \)
At \( t = -4 \), \( a(-4) = \frac{50}{1^3} = 50 \text{ m/s}^2 \)
This shows that acceleration varies significantly at different points in time, depending on how the velocity changes at those points.