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

a. Displacement: -20 m, Average Velocity: -5 m/s. b. Speeds: 25 m/s (at -4), 1 m/s (at 0); Accelerations: 50 m/s² (at -4), 0.4 m/s² (at 0). c. The body does not change direction.

Step 1: Calculate Displacement

The displacement of the body is the change in position from the start to the end of the interval. Compute the position at the start, \( t = -4 \), and at the end, \( t = 0 \): \[ s(-4) = \frac{25}{-4+5} = 25 \] \[ s(0) = \frac{25}{0+5} = 5 \] Displacement, \( \Delta s = s(0) - s(-4) = 5 - 25 = -20 \text{ meters} \).

Step 2: Calculate Average Velocity

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

Step 3: Find Speed and Acceleration at Endpoints

Speed is the magnitude of velocity. To find velocity and acceleration, differentiate \( s(t) \):Velocity: \( v(t) = \frac{d}{dt} \left( \frac{25}{t+5} \right) = -\frac{25}{(t+5)^2} \) Acceleration: \( a(t) = \frac{d}{dt} \left( -\frac{25}{(t+5)^2} \right) = \frac{50}{(t+5)^3} \)At \( t = -4 \): \[ v(-4) = -\frac{25}{1^2} = -25 \text{ m/s}, \quad a(-4) = \frac{50}{1^3} = 50 \text{ m/s}^2 \]. Speed is \(|v(-4)| = 25 \text{ m/s}\).At \( t = 0 \): \[ v(0) = -\frac{25}{5^2} = -1 \text{ m/s}, \quad a(0) = \frac{50}{5^3} = \frac{2}{5} \text{ m/s}^2 \]. Speed is \(|v(0)| = 1 \text{ m/s} \).

Step 4: Determine When Body Changes Direction

The body changes direction when the velocity changes sign. Set \( v(t) = 0 \) and solve for \( t \):\[ -\frac{25}{(t + 5)^2} = 0 \] The equation never equals zero because the numerator \(-25\) cannot be zero. Thus, the velocity does not change sign and the body does not change direction in the given interval.

Key Concepts

displacement

Displacement is a term used in physics to describe the change in position of a body. It involves determining how far the object moves from its initial position to its final position over a specific time period. In calculus problems like this, it is computed by evaluating the function that describes motion at the beginning and ending of the interval.

To solve for displacement, we calculated the position of the body at the start of the interval (\( t = -4 \)) and at the end (\( t = 0 \)). Using the function \( s(t) = \frac{25}{t+5} \), we find:

• Start position: \( s(-4) = 25 \)
• End position: \( s(0) = 5 \)

The displacement, \( \Delta s \), is then calculated by subtracting the start position from the end position: \( \Delta s = 5 - 25 = -20 \) meters. This negative value indicates the body has moved backward in the given interval.

average velocity

Average velocity is a measure of the total change in position (displacement) divided by the time interval during which the change occurred. It essentially gives an average rate of motion, considering how much ground was covered and how long it took.

In this problem, the average velocity, \( v_{avg} \), is found using the formula \( v_{avg} = \frac{\Delta s}{\Delta t} \). Given that the displacement \( \Delta s = -20 \) meters and the time interval spans from \( t = -4 \) to \( t = 0 \) (a duration of 4 seconds), we compute:

• \( v_{avg} = \frac{-20}{4} = -5 \text{ m/s} \)

The negative sign of the average velocity shows that the body's motion is directed backwards over the interval.

acceleration

Acceleration is the rate at which an object changes its velocity. In calculus, we find acceleration by differentiating the velocity function with respect to time. It tells how quickly or slowly a body speeds up or slows down.

In our exercise, the velocity function is derived by differentiating the position function. Given \( s(t) = \frac{25}{t+5} \), the velocity \( v(t) \) and acceleration \( a(t) \) are determined as:

• Velocity: \( v(t) = -\frac{25}{(t+5)^2} \)
• Acceleration: \( a(t) = \frac{50}{(t+5)^3} \)

Examining the endpoints:

• At \( t = -4 \), acceleration is \( 50 \text{ m/s}^2 \)
• At \( t = 0 \), acceleration is \( \frac{2}{5} \text{ m/s}^2 \)

These values indicate how the velocity is increasing or decreasing at those moments.

direction change

A change in direction occurs when an object reverses its path of motion, which means its velocity changes sign (from positive to negative or vice versa).

For the body in our calculus problem to change direction, its velocity function \( v(t) \) must equal zero at some point in the interval. We determine if and where this occurs by setting the velocity function to zero and solving for \( t \). However, since \( v(t) = -\frac{25}{(t+5)^2} \), it is clear that this expression cannot be zero — the numerator is a constant \(-25\) and thus cannot satisfy the condition of equating to zero.

Consequently, the body's velocity does not cross zero in the given interval, indicating there is no change in direction. It maintains the same direction throughout the period.