Problem 1 - University Calculus: Early Transcendentals

Question

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? $$s=t^{2}-3 t+2, \quad 0 \leq t \leq 2$$

Step-by-Step Solution

Answer

a. Displacement is -2 m; average velocity is -1 m/s. b. Speeds: 3 m/s (0 s), 1 m/s (2 s); acceleration: 2 m/s². c. Changes direction at t = 1.5 s.

Step 1: Expression of Displacement

The displacement of the body over a time interval can be calculated using the change in position, given by \( \Delta s = s(t_2) - s(t_1) \). Given the function \( s(t) = t^2 - 3t + 2 \) and the interval \( 0 \le t \le 2 \), find \( s(2) \) and \( s(0) \).

Step 2: Calculate Positions

Calculate \( s(2) \) and \( s(0) \) using the function \( s(t) = t^2 - 3t + 2 \). \[ s(0) = 0^2 - 3 \cdot 0 + 2 = 2 \] \[ s(2) = 2^2 - 3 \cdot 2 + 2 = 0 \].

Step 3: Calculate Displacement

The displacement is the difference between the final and the initial positions: \[ \Delta s = s(2) - s(0) = 0 - 2 = -2 \text{ meters} \].

Step 4: Expression of Average Velocity

Average velocity \( v_{avg} \) over the interval is given by \( v_{avg} = \frac{\Delta s}{\Delta t} \), where the interval \( \Delta t = 2 - 0 \).

Step 5: Calculate Average Velocity

Using the displacement \( \Delta s = -2 \) meters and interval \( \Delta t = 2 \) seconds, calculate the average velocity: \[ v_{avg} = \frac{-2}{2} = -1 \text{ m/s} \].

Step 6: Expression of Speed and Acceleration

Speed is the absolute value of velocity. To find the velocity as a function of time, differentiate \( s(t) \): \( v(t) = \frac{ds}{dt} = 2t - 3 \). To find acceleration, differentiate velocity: \( a(t) = \frac{dv}{dt} = 2 \).

Step 7: Calculate Velocity at Endpoints

Find velocity at the endpoints \( t = 0 \) and \( t = 2 \): At \( t = 0 \): \[ v(0) = 2 \cdot 0 - 3 = -3 \text{ m/s} \] At \( t = 2 \): \[ v(2) = 2 \cdot 2 - 3 = 1 \text{ m/s} \].

Step 8: Calculate Speed at Endpoints

Since speed is the absolute value of velocity: Speed at \( t = 0 \): \[ |v(0)| = |-3| = 3 \text{ m/s} \] Speed at \( t = 2 \): \[ |v(2)| = |1| = 1 \text{ m/s} \].

Step 9: Calculate Acceleration at Endpoints

Since acceleration is constant \( a(t) = 2 \text{ m/s}^2 \) everywhere along the interval, it is the same at both endpoints \( t = 0 \) and \( t = 2 \).

Step 10: Determine Change in Direction

The body changes direction if the velocity changes sign. Set \( v(t) = 0 \) to solve for \( t \): \( v(t) = 2t - 3 \) \( 2t - 3 = 0 \) \( 2t = 3 \) \( t = \frac{3}{2} \). Since \( t = \frac{3}{2} \) falls within the interval \( [0, 2] \), the body changes direction at \( t = \frac{3}{2} \) seconds.

Key Concepts

Displacement

Displacement refers to the change in position of an object during a given time interval. It is a vector quantity, meaning it has both magnitude and direction. For our specific exercise, the displacement is computed through the expression \[ \Delta s = s(t_2) - s(t_1) \] where \(s(t)\) is a function representing the position of the object with respect to time \(t\). In this case, the position function is \(s(t) = t^2 - 3t + 2\). When calculated over the interval \(0 \le t \le 2\) seconds:

• Initial position, \(s(0)\): \(0^2 - 3 \times 0 + 2 = 2\) meters
• Final position, \(s(2)\): \(2^2 - 3 \times 2 + 2 = 0\) meters

Hence, the displacement is calculated as \(\Delta s = 0 - 2 = -2\) meters. The negative sign indicates the body has moved 2 meters backwards in the direction of motion.

Average Velocity

Average velocity helps determine the "overall" speed and direction of an object's motion over a specific period. It differs from instantaneous velocity, which measures speed at a particular moment. The formula to calculate average velocity is\[ v_{avg} = \frac{\Delta s}{\Delta t} \]where \( \Delta s\) is displacement and \( \Delta t\) is the time interval duration. For our exercise, with a displacement of -2 meters over 2 seconds:

• The time interval \( \Delta t = 2 - 0 = 2 \) seconds
• The average velocity: \( v_{avg} = \frac{-2}{2} = -1 \text{ m/s} \)

The negative average velocity indicates that the motion over the entire interval is in the opposite direction to positive displacement.

Acceleration

Acceleration measures how quickly an object's velocity changes over time. It can be constant or variable, and is a vector quantity like displacement and velocity, meaning it carries both a magnitude and a direction. For an object moving according to the position function \(s(t) = t^2 - 3t + 2\), the velocity function, derived by differentiating the position function, is:\[ v(t) = \frac{ds}{dt} = 2t - 3 \]By differentiating the velocity function further, we obtain the acceleration function:\[ a(t) = \frac{dv}{dt} = 2 \text{ m/s}^2 \]In this case, the acceleration is constant throughout the interval \([0, 2]\) seconds, indicating a uniform change in velocity.

Direction Change

A body changes direction when its velocity changes sign. This occurs when the object transitions from moving in one direction to the opposite. To determine where this takes place involves solving when the velocity function equals zero. Given the velocity function\[ v(t) = 2t - 3 \]solve for \(t\) by setting\[ 2t - 3 = 0 \]Rearrange to find:

• \( 2t = 3 \)
• \( t = \frac{3}{2} \) seconds

Since \( t = \frac{3}{2} \approx 1.5\) is within our interval \(0 \leq t \leq 2\), it signifies the point at which the body changes its direction.