Problem 4 - 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? $$s=\left(t^{4} / 4\right)-t^{3}+t^{2}, \quad 0 \leq t \leq 3$$

Step-by-Step Solution

Answer

a) Displacement: \(9/4\) m, Avg. Velocity: \(3/4\) m/s. b) Speed: 0 m/s (\(t=0\)), 6 m/s (\(t=3\)); Accel.: 2 m/s² (\(t=0\)), 11 m/s² (\(t=3\)). c) Changes direction at \(t=1\) and \(t=2\).

Step 1: Calculate Displacement over the Interval

To find the displacement of the body over the interval \([0, 3]\), we need to calculate the change in position, \(s\), from the starting time \(t = 0\) to the ending time \(t = 3\). Displacement is found by subtracting the initial position from the final position: \[ s_{0} = f(0) = \frac{0^4}{4} - 0^3 + 0^2 = 0 \] \[ s_{3} = f(3) = \frac{3^4}{4} - 3^3 + 3^2 = \frac{81}{4} - 27 + 9 = \frac{9}{4} \] Therefore, the displacement is: \[ \text{Displacement} = s_{3} - s_{0} = \frac{9}{4} - 0 = \frac{9}{4} \text{ meters} \]

Step 2: Calculate Average Velocity over the Interval

Average velocity is defined as the total displacement divided by the total time taken. Using the displacement found in Step 1 and the time interval \([0, 3]\), the average velocity is: \[ \text{Average Velocity} = \frac{\text{Displacement}}{\Delta t} = \frac{\frac{9}{4}}{3 - 0} = \frac{9}{12} = \frac{3}{4} \text{ meters per second} \]

Step 3: Find Instantaneous Speed at Endpoints

The instantaneous speed is the absolute value of the velocity. To find the velocity, take the derivative of the position function \(s(t)\): \[ v(t) = s'(t) = \frac{d}{dt}\left(\frac{t^4}{4} - t^3 + t^2\right) = t^3 - 3t^2 + 2t \]Evaluate the velocity at \(t = 0\) and \(t = 3\):\[ v(0) = 0^3 - 3(0)^2 + 2(0) = 0 \]\[ v(3) = 3^3 - 3(3)^2 + 2(3) = 27 - 27 + 6 = 6 \]Thus, the speeds are:- Speed at \(t=0\): \( |v(0)| = 0 \text{ m/s} \)- Speed at \(t=3\): \( |v(3)| = 6 \text{ m/s} \)

Step 4: Find Acceleration at Endpoints

Acceleration is the derivative of the velocity function. Differentiate \(v(t)\):\[ a(t) = v'(t) = \frac{d}{dt}(t^3 - 3t^2 + 2t) = 3t^2 - 6t + 2 \]Calculate acceleration at \(t = 0\) and \(t = 3\):\[ a(0) = 3(0)^2 - 6(0) + 2 = 2 \]\[ a(3) = 3(3)^2 - 6(3) + 2 = 27 - 18 + 2 = 11 \]Thus, the accelerations are:- Acceleration at \(t=0\): \(2 \text{ m/s}^2\)- Acceleration at \(t=3\): \(11 \text{ m/s}^2\)

Step 5: Check for Change of Direction

The body changes direction where the velocity is zero and changes sign. Solve \(v(t) = 0\):\[ t^3 - 3t^2 + 2t = t(t^2 - 3t + 2) = t(t-1)(t-2) = 0 \]This gives solutions \(t=0, t=1, t=2\). Evaluate the signs of \(v(t)\) around these points:- Interval \((0, 1)\): Test with \(t = 0.5\), \(v(0.5) = (0.5)^3 - 3(0.5)^2 + 2(0.5) = 0.125 - 0.75 + 1 = 0.375\) (positive)- Interval \((1, 2)\): Test with \(t = 1.5\), \(v(1.5) = (1.5)^3 - 3(1.5)^2 + 2(1.5) = 3.375 - 6.75 + 3 = -0.375\) (negative)- Interval \((2, 3)\): Test with \(t = 2.5\), \(v(2.5) = (2.5)^3 - 3(2.5)^2 + 2(2.5) = 15.625 - 18.75 + 5 = 1.875\) (positive)The body changes direction at \(t=1\) and \(t=2\).

Key Concepts

Displacement

Displacement is an important concept in calculus and physics, representing the change in position of an object over a given time interval. Simply put, it tells you how far out of place an object is after moving. To find displacement, you subtract the initial position from the final position. In formulas, this is seen as \[ \text{Displacement} = s(t_\text{final}) - s(t_\text{initial}) \]. In our exercise example, the displacement over the interval from time \(t = 0\) to \(t = 3\) seconds is calculated using the position function \(s(t)\). We first find \(s(0)\) and \(s(3)\), which are the positions at the start and end of the interval, respectively. Calculating these gives us:

• \(s(0) = 0\)
• \(s(3) = \frac{9}{4} \text{ meters}\)

Thus, the displacement is \(\frac{9}{4}\text{ meters}\). It is crucial to note that displacement is direction sensitive and can be negative or positive.

Average Velocity

Average velocity is defined as the total displacement divided by the total time taken during the interval. This measurement provides the rate at which an object changes its position. Mathematically, average velocity is given by:\[ \text{Average Velocity} = \frac{\text{Displacement}}{\Delta t} \]where \(\Delta t\) is the change in time.For our interval between \(t = 0\) and \(t = 3\), using our calculated displacement of \(\frac{9}{4}\text{ meters}\), the formula becomes:

• \(\text{Average Velocity} = \frac{\frac{9}{4}}{3 - 0} = \frac{9}{12} = \frac{3}{4} \text{ meters per second}\)

Average velocity provides a useful summary of motion over a period. It doesn't, however, detail changes during the interval.

Acceleration

Acceleration measures how the velocity of an object changes with time. Unlike average velocity, acceleration looks at the rate of change of speed or velocity, making it vital for understanding dynamic motion.To determine acceleration, we take the derivative of the velocity function with respect to time. The velocity function, the derivative of the position function \(s(t)\), in our case is:\[ v(t) = t^3 - 3t^2 + 2t \]The acceleration function is then\[ a(t) = v'(t) = 3t^2 - 6t + 2 \].We calculate acceleration at endpoints \(t = 0\) and \(t = 3\) as follows:

• \(a(0) = 2 \text{ m/s}^2\)
• \(a(3) = 11 \text{ m/s}^2\)

These values tell us how rapidly speeds are changing at these moments, indicating how fast the velocity is increasing or decreasing.

Direction Change

A change in direction occurs when an object stops moving in one direction and begins moving in the opposite direction. In terms of calculus, a direction change happens when the velocity of the object is zero and then changes sign (positive to negative or negative to positive).To identify these moments mathematically, we set the velocity function \(v(t)\) to zero and solve for \(t\).\[ t(t - 1)(t - 2) = 0 \]solving this gives solutions:

• \(t = 0\)
• \(t = 1\)
• \(t = 2\)

By evaluating the velocity's sign before and after these times, we determine that the direction of motion changes at \(t = 1\) and \(t = 2\). Understanding when and where an object changes direction is crucial in analyzing an object's complete motion path.