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

Step-by-Step Solution

Answer

The displacement is -9 meters, average velocity is -3 m/s, speed changes from 3 m/s to 12 m/s, acceleration is 6 m/s² and -12 m/s², and direction changes at t = 1.

Step 1: Identify the Time Interval and Function

The function provided is \(s(t) = -t^3 + 3t^2 - 3t\). We are asked to analyze this function over the interval \(0 \leq t \leq 3\).

Step 2: Calculate the Displacement

Displacement is the change in position over the time interval. We calculate the position at the endpoints of the interval: \[s(3) = -(3)^3 + 3(3)^2 - 3(3) = -27 + 27 - 9 = -9\] \[s(0) = -(0)^3 + 3(0)^2 - 3(0) = 0\]Displacement \( = s(3) - s(0) = -9 - 0 = -9\, \text{meters}\).

Step 3: Calculate the Average Velocity

Average velocity is the total displacement divided by the total time: \[\text{Average Velocity} = \frac{-9}{3 - 0} = \frac{-9}{3} = -3\, \text{m/s}\].

Step 4: Find Speed and Acceleration at Endpoints

Speed is the magnitude of velocity. We find the velocity function by differentiating the position function:\[v(t) = \frac{ds}{dt} = -3t^2 + 6t - 3\].Find speed at \(t = 0\):\[v(0) = -3(0)^2 + 6(0) - 3 = -3\], hence speed is \(3\,\text{m/s}\).Find speed at \(t = 3\):\[v(3) = -3(3)^2 + 6(3) - 3 = -27 + 18 - 3 = -12\], so speed is \(12\,\text{m/s}\).Acceleration is given by the derivative of velocity:\[a(t) = \frac{dv}{dt} = -6t + 6\].Find acceleration at \(t = 0\):\[a(0) = -6(0) + 6 = 6\, \text{m/s}^2\].Find acceleration at \(t = 3\):\[a(3) = -6(3) + 6 = -18 + 6 = -12\, \text{m/s}^2\].

Step 5: Determine if Body Changes Direction

The body changes direction if the velocity changes sign. Solve for when \(v(t) = 0\):\[-3t^2 + 6t - 3 = 0\].Factoring, we get:\[0 = -3(t^2 - 2t + 1) = -3(t-1)^2\].This gives \(t = 1\). Since \(v(t)\) changes sign, the body changes direction at \(t = 1\).

Key Concepts

Acceleration

Acceleration tells us how quickly the velocity of an object is changing. It's like a measure of how fast the speed is speeding up or slowing down. Mathematically, acceleration is the derivative of the velocity function. This means we are looking at how the velocity changes with respect to time.
In our exercise, we start with the velocity equation given by differentiating the position function. The position function is: - \(s(t) = -t^3 + 3t^2 - 3t\). When we differentiate this function, we obtain the velocity function \(v(t) = -3t^2 + 6t - 3\). To find the acceleration, we differentiate once more:- \(a(t) = \frac{dv}{dt} = -6t + 6\). Acceleration gives us critical information:

• At \(t = 0\), the acceleration is \(6\, \text{m/s}^2\), indicating that the velocity is increasing initially.
• At \(t = 3\), the acceleration is \(-12\, \text{m/s}^2\), signifying a decrease in velocity towards the end of the interval.

Speed

Speed is like the absolute value of velocity. It tells us how fast an object is moving, regardless of the direction. While velocity can be negative (indicating direction), speed is a non-negative quantity that solely focuses on magnitude.
To find speed at specific points:- Calculate the velocity using the velocity function \(v(t)\), which we derived as \(-3t^2 + 6t - 3\). For our exercise:

• At \(t = 0\), velocity \(v(0) = -3\). Therefore, speed, which is the absolute value, is \(3 \text{ m/s}\).
• At \(t = 3\), velocity \(v(3) = -12\). Hence, speed is \(12 \text{ m/s}\).

Speed provides a clear picture of how fast the object actually travels without worrying about which direction it goes.

Direction Change

The concept of direction change is closely related to changes in the sign of velocity. When an object's velocity changes from positive to negative or vice versa, it means the object has changed direction.
For our given problem:- We identify directions by solving \(v(t) = 0\). The velocity function is \(-3t^2 + 6t - 3\). Solving for \(v(t) = 0\) gives us:- \[-3(t - 1)^2 = 0\] This results in \(t = 1\). Because the velocity crosses zero, the body changes direction at this point.

• Before \(t = 1\), the velocity was in one direction.
• After \(t = 1\), it shifted to the opposite.

The notion of velocity's sign change is crucial in identifying these shifts and understanding the trajectory of a moving object.