Problem 2 - 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? $$s=6 t-t^{2}, \quad 0 \leq t \leq 6$$

Step-by-Step Solution

Answer

a) Displacement: 0 m, Average Velocity: 0 m/s. b) Speed at t=0: 6 m/s, at t=6: 6 m/s; Acceleration: -2 m/s². c) Direction change at t=3.

Step 1: Calculate Displacement

To find the displacement of the body over the interval, calculate the change in position (or function value) by evaluating the position function at the endpoints. The displacement is given by \( s(f(6)) - s(f(0)) \). First calculate:\[ s(0) = 6(0) - (0^2) = 0 \]\[ s(6) = 6(6) - (6^2) = 36 - 36 = 0 \]The displacement over the interval \([0, 6]\) is:\[ 0 - 0 = 0 \] meters.

Step 2: Calculate Average Velocity

The average velocity is the displacement divided by the time interval. Since the displacement is 0, the average velocity is:\[ \text{Average Velocity} = \frac{0}{6 - 0} = 0 \, \text{m/s} \]

Step 3: Find Velocity Function

The velocity function is the derivative of the position function. Differentiate \( s = 6t - t^2 \):\[ v(t) = \frac{d}{dt}(6t - t^2) = 6 - 2t \]This gives the velocity of the body at any time \(t\).

Step 4: Calculate Speed and Acceleration at Endpoints

Acceleration is the derivative of velocity. Differentiate \( v(t) = 6 - 2t \) to get acceleration:\[ a(t) = \frac{d}{dt}(6 - 2t) = -2 \]Calculate both velocity and speed at \( t = 0 \) and \( t = 6 \):- Velocity at \( t = 0 \): \[ v(0) = 6 - 2(0) = 6 \, \text{m/s} \] Speed is \( 6 \, \text{m/s} \). Acceleration is \( -2 \, \text{m/s}^2 \).- Velocity at \( t = 6 \): \[ v(6) = 6 - 2(6) = 6 - 12 = -6 \, \text{m/s} \] Speed is \( 6 \, \text{m/s} \). Acceleration is \( -2 \, \text{m/s}^2 \).

Step 5: Determine If the Body Changes Direction

The body changes direction when the velocity changes sign. Solve for when \( v(t) = 0 \):\[ 6 - 2t = 0 \]\[ 2t = 6 \]\[ t = 3 \]At \( t = 3 \), the velocity changes from positive to negative, indicating a change in direction.

Key Concepts

Displacement

Displacement refers to the change in position of an object along a line. It doesn't matter how the object got there, just the start and end points. For a position function like \( s(t) = 6t - t^2 \), finding displacement involves calculating the position at the start and end of the interval, and then finding the difference. In this case, the positions are calculated at \( t = 0 \) and \( t = 6 \). Both positions yield the same value: 0 meters. Displacement, therefore, is \( 0 - 0 = 0 \) meters. Even if the object moved during the time, if it ends where it started, the displacement is zero.
This might seem counterintuitive for something like a journey with a return trip, but displacement only considers the final positions, not the path taken.

Average Velocity

Average velocity is a measure of the overall change in position (displacement) over a specific time interval. It tells us how fast an object is moving, *on average* from one point in time to another. To find the average velocity, divide the total displacement by the time taken. For the interval from \( t = 0 \) to \( t = 6 \), since the displacement is 0 meters, the average velocity is also 0 m/s.
This highlights that although the object might have been moving, since it ends at the same place it started within the specified interval, its average velocity is zero. This is different from speed because average speed would consider the total path traveled.

Acceleration

Acceleration describes how the velocity of an object changes over time. You find acceleration by taking the derivative of the velocity function. In the case where velocity \( v(t) = 6 - 2t \), the derivative is \( a(t) = -2 \), which tells us that the object has a constant acceleration of \(-2\, \text{m/s}^2\).

• At \( t = 0 \), the initial velocity is 6 m/s.
• As time progresses, the velocity decreases by 2 m/s every second.
• This consistent change in velocity illustrates the object's steady deceleration.

Acceleration can either increase or decrease speed depending on its direction; in this scenario, it indicates a constant slowing down.

Velocity Function

The velocity function is derived by differentiating the position function, \( s(t) = 6t - t^2 \). Differentiation provides \( v(t) = 6 - 2t \), which represents the velocity at any moment \( t \).
The velocity function tells us:

• Initially, at \( t=0 \), the velocity is 6 m/s.
• The velocity decreases over time due to the constant negative acceleration.
• At \( t=3 \), velocity becomes zero, indicating a momentary stop and the point at which direction changes.
• Continuing beyond \( t=3 \), the velocity becomes negative, showing the object reverses direction.

The velocity function is critical in understanding not just how fast an object moves but also in predicting how it will change direction and speed.