Problem 9 - Calculus Early Transcendentals

Question

Assume \(t\) is time measured in seconds and velocities have units of \(\mathrm{m} / \mathrm{s}\). a. Graph the velocity function over the given interval. Then determine when the motion is in the positive direction and when it is in the negative direction. b. Find the displacement over the given interval. c. Find the distance traveled over the given interval. $$v(t)=t^{2}-6 t+8 ; 0 \leq t \leq 5$$

Step-by-Step Solution

Answer

Question: Determine when the motion is in the positive and negative directions, the displacement, and the distance traveled over the interval \(0 \leq t \leq 5\) for the velocity function \(v(t) = t^2 - 6t + 8\). Answer: The motion is in the positive direction for \(0 \leq t < 2\) and \(4 \leq t \leq 5\), and in the negative direction for \(2 \leq t < 4\). The displacement over the interval is \(\frac{10}{3}\ \mathrm{m}\), and the distance traveled is \(\frac{70}{3}\ \mathrm{m}\).

Step 1: Graph the velocity function

To graph the function \(v(t) = t^2 - 6t + 8\), we can use a graphing calculator or software. By looking at the graph, we can visually determine when the motion is positive or negative.

Step 2: Finding when motion is positive or negative

Now, we need to find when the motion is positive or negative by analyzing the function. To do this, we'll find the roots of the function and analyze the sign of \(v(t)\) between those roots. Roots are the points when the function is equal to zero, i.e., \(v(t)=0\). Solve the quadratic equation: \(t^2-6t+8=0\). We find the roots are \(t = 2\) and \(t = 4\). Now let's test the intervals \((0,2)\), \((2,4)\), and \((4,5)\). 1. For \(0 0\) (Positive motion) 2. For \(2 0\) (Positive motion) So, motion is in positive direction for \(0 \leq t < 2\) and \(4 \leq t \leq 5\), and motion is in negative direction for \(2 \leq t < 4\). #b. Finding the displacement over the given interval#

Step 3: Calculate the displacement

Displacement can be found by integrating the velocity function over the given interval. To find the displacement, we'll compute the definite integral: $$\int_{0}^{5}(t^2 - 6t + 8)dt$$ Let's integrate and then evaluate: $$\left[\frac{t^3}{3} - 3t^2 + 8t\right]_0^5 = \frac{125}{3} - 75 + 40 = \frac{10}{3}$$ So, the displacement is \(\frac{10}{3}\ \mathrm{m}\). #c. Finding the distance traveled over the given interval#

Step 4: Calculate the distance traveled

To find the distance traveled, we'll need to integrate the absolute value of the velocity function over each of the subintervals we found in Step 2. $$\int_0^2(t^2-6t+8)dt + \int_2^4(-(t^2-6t+8))dt + \int_4^5(t^2-6t+8)dt$$ First integral: $$\int_0^2(t^2-6t+8)dt = \left[\frac{t^3}{3} - 3t^2 + 8t\right]_0^2 = \frac{8}{3}$$ Second integral: $$\int_2^4(-(t^2-6t+8))dt = -\left[\frac{t^3}{3} - 3t^2 + 8t\right]_2^4 = \frac{40}{3}$$ Third integral: $$\int_4^5(t^2-6t+8)dt = \left[\frac{t^3}{3} - 3t^2 + 8t\right]_4^5 = \frac{22}{3}$$ Now we add up the results: $$\frac{8}{3} + \frac{40}{3} + \frac{22}{3} = \frac{70}{3}$$ So, the distance traveled over the interval is \(\frac{70}{3}\ \mathrm{m}\).

Key Concepts

Positive and Negative Motion

In a velocity function, motion can be either positive or negative, indicating the direction of movement. This concept is crucial when analyzing physics problems involving velocity over time. A velocity function, like the quadratic equation in our exercise, can be graphed to visually interpret these directions. Positive motion occurs when the velocity function yields a positive value, indicating movement in a forward direction. Conversely, negative motion arises when the function produces a negative value, pointing to a reverse direction. To find these intervals within a given time span, it is essential to determine where the velocity function equals zero, as this marks the transition between positive and negative motion. These points, called roots, can be found by solving the equation set to zero. For the function \(v(t)=t^{2}-6t+8\), the roots are determined to be \(t = 2\) and \(t = 4\).

• For \(t\) in the range of \(0 \leq t < 2\), the velocity is positive.
• For \(2 \leq t < 4\), the velocity turns negative.
• For \(4 \leq t \leq 5\), the motion becomes positive again.

Displacement

Displacement is a vector quantity that represents the overall change in position of an object. It considers the direction of motion and provides the net change from the starting point to the endpoint. In our exercise, displacement is calculated by finding the definite integral of the velocity function over the specific time interval. This integral gives the area under the velocity-time graph, providing a measure of the total positional change. For the velocity function \(v(t) = t^2 - 6t + 8\) over the interval from \(t = 0\) to \(t = 5\), we integrate to get:\[\int_{0}^{5}(t^2 - 6t + 8)\,dt\]Evaluating this integral, we find the displacement to be \(\frac{10}{3} \text{ m}\). This value indicates the net change in position of the object along its path during the time considered.

Distance Traveled

The distance traveled differs from displacement as it's a scalar quantity representing the total path length an object covers, regardless of its direction. This means we take into account all segments of motion, positive or negative, and sum them up to find the complete journey length. To compute the distance traveled, especially when direction changes, it is necessary to split the velocity function into intervals where the sign of the function is constant. Each interval must be considered separately to correctly account for absolute values, turning any negative values into positives.For the exercise’s velocity function \(v(t) = t^{2} - 6t + 8\) on the interval \(0 \leq t \leq 5\), we separate it based on positive and negative sections by integrating:

• \(\int_0^2(t^2-6t+8)\,dt\)
• \(\int_2^4(-(t^2-6t+8))\,dt\)
• \(\int_4^5(t^2-6t+8)\,dt\)

The respective evaluated results are added up, leading to a total distance traveled of \(\frac{70}{3} \text{ m}\). This comprehensive calculation gives a true picture of the entire path covered.