Problem 1 - 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? $$ s=t^{2}-3 t+2, \quad 0 \leq t \leq 2 $$

Step-by-Step Solution

Answer

Displacement: -2 meters; Average velocity: -1 m/s; Speed at endpoints: 3 m/s and 1 m/s; Acceleration: 2 m/s²; Direction change at t = 1.5s.

Step 1: Calculate displacement

The displacement of a body over a time interval \[ [t_1, t_2] \]is given by the difference in position at the endpoints:\[ \Delta s = f(t_2) - f(t_1) \]Given \[ s = t^2 - 3t + 2 \]and the interval \[ 0 \leq t \leq 2 \], calculate the positions at \( t = 0 \) and \( t = 2 \):\[ f(0) = (0)^2 - 3(0) + 2 = 2 \]\[ f(2) = (2)^2 - 3(2) + 2 = 4 - 6 + 2 = 0 \]Thus, \[ \Delta s = 0 - 2 = -2 \] meters.

Step 2: Find average velocity

The average velocity \( v_{avg} \) over the time interval \( [t_1, t_2] \) is given by:\[ v_{avg} = \frac{\Delta s}{\Delta t} \]Plug in the values for displacement and time interval:\[ v_{avg} = \frac{-2}{2-0} = -1 \text{ m/s} \]

Step 3: Calculate velocity

Velocity \( v(t) \) is the first derivative of position \( s \). Differentiate \( s(t) = t^2 - 3t + 2 \):\[ v(t) = \frac{d}{dt}(t^2 - 3t + 2) = 2t - 3 \]

Step 4: Calculate speed and acceleration at endpoints

Speed is the absolute value of velocity. First, calculate velocity at the endpoints \( t = 0 \) and \( t = 2 \):At \( t = 0 \):\[ v(0) = 2(0) - 3 = -3 \text{ m/s} \]Speed \( = |-3| = 3 \text{ m/s} \)At \( t = 2 \):\[ v(2) = 2(2) - 3 = 1 \text{ m/s} \]Speed \( = |1| = 1 \text{ m/s} \)Acceleration \( a(t) \) is the derivative of velocity. Differentiate \( v(t) = 2t - 3 \):\[ a(t) = \frac{d}{dt}(2t - 3) = 2 \]Acceleration is constant at \( 2 \text{ m/s}^2 \) for both \( t = 0 \) and \( t = 2 \).

Step 5: Check for direction change

A change in direction occurs when velocity changes sign. Find when \( v(t) = 0 \):\[ 2t - 3 = 0 \]\[ 2t = 3 \]\[ t = \frac{3}{2} \]Since \( \frac{3}{2} = 1.5 \) is within the interval \( 0 \leq t \leq 2 \), the body changes direction at \( t = 1.5 \) seconds.

Key Concepts

Average Velocity

In calculus, average velocity helps us understand the overall movement of a body over a time period. It's the mid-point value of speed that encapsulates the entire journey between two points in time.
For a given position function expressed as \( s(t) = t^2 - 3t + 2 \) with time interval \( 0 \leq t \leq 2 \), average velocity is calculated using the formula:

• \( v_{avg} = \frac{\Delta s}{\Delta t} \)

To find \( \Delta s \), which is displacement, we evaluate \( s(t) \) at \( t = 0 \) and \( t = 2 \):
- At \( t = 0 \), \( s(0) = 2 \) meters.
- At \( t = 2 \), \( s(2) = 0 \) meters.
Thus, the displacement \( \Delta s \) is \( 0 - 2 = -2 \) meters.
The time interval \( \Delta t \) is \( 2 - 0 = 2 \) seconds. Thus, the average velocity is:
\( v_{avg} = \frac{-2}{2} = -1 \text{ m/s} \). This negative value indicates backward motion on the coordinate line.

Speed and Acceleration

Speed and acceleration are key concepts that describe an object's motion relative to time.
Speed is simply the absolute value of velocity, indicating how fast an object moves regardless of direction.
Meanwhile, acceleration tells us how quickly the velocity is changing.
- **Velocity Function:** To understand speed, first calculate velocity, which is the derivative of the position function:
\( v(t) = \frac{d}{dt}(t^2 - 3t + 2) = 2t - 3 \).

• At \( t = 0 \), \( v(0) = -3 \text{ m/s} \) yielding a speed of \( 3 \text{ m/s} \).
• At \( t = 2 \), \( v(2) = 1 \text{ m/s} \) giving a speed of \( 1 \text{ m/s} \).

- **Acceleration:** It's the derivative of the velocity function:
\( a(t) = \frac{d}{dt}(2t - 3) = 2 \text{ m/s}^2 \).
This constant acceleration means that the object's velocity increases by \( 2 \text{ m/s} \) every second between \( t = 0 \) and \( t = 2 \).
It gives us insight into how the motion is changing over time.

Direction Change in Motion

A change of direction in motion is closely related to when velocity transitions from positive to negative, or vice versa. To determine these points, we identify when velocity equals zero.
The velocity equation we derived is:
\( v(t) = 2t - 3 \).
To find when \( v(t) = 0 \):

• Solve \( 2t - 3 = 0 \).
• We find \( t = \frac{3}{2} = 1.5 \) seconds.

Since \( t = 1.5 \) seconds lies within our defined interval \( 0 \leq t \leq 2 \), it indicates a moment where the body's direction shifts.
This understanding of changes in motion is crucial in various applications, such as physics and engineering, ensuring we appreciate how speed and direction entwine to define movement.