Problem 11
Question
Assume \(t\) is time measured in seconds and velocities have units of \(m / s\) 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. $$v(t)=t^{2}-6 t+8 \text { on } 0 \leq t \leq 5$$
Step-by-Step Solution
Verified Answer
a. The motion is positive on the intervals \(0\leq t<2\) and \(4
1Step 1: Find when the motion is positive and negative
Solve \(v(t)=0\) to find when the object is stationary.
\(t^2 - 6t + 8 = 0\) Factor the equation:
\((t-2)(t-4)=0\)
So, \(t=2\) and \(t=4\)
Now, we'll examine the motion in the three intervals: \(0\leq t<2\), \(20\)
For \(20\)
Therefore, the motion is positive on \(0\leq t<2\) and \(4
2Step 2: Find the displacement (net change in position)
To find the displacement, integrate the velocity function from \(t=0\) to \(t=5\):
$$\int_0^5 (t^2 - 6t + 8) dt$$
Now, integrate the function to get the displacement:
$$\left[\frac{t^3}{3} - 3t^2 + 8t\right]_0^5 = \left(\frac{125}{3} - 75 + 40\right) - \left(0 - 0 + 0\right) = \frac{10}{3}.$$
So, the displacement is \(\frac{10}{3} m\).
3Step 3: Find the distance traveled
To find the total distance traveled, first integrate the absolute value of the velocity function on each interval where the motion is either positive or negative.
For \(0\leq t<2\) and \(4
Key Concepts
DisplacementDistance TraveledInterval Analysis
Displacement
Displacement in physics refers to the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction. In this problem, the displacement is calculated by integrating the velocity function over the given interval of time, which is from 0 to 5 seconds.
The velocity function given is \( v(t) = t^2 - 6t + 8 \). To find the displacement, we integrate this function:
\[ \int_0^5 (t^2 - 6t + 8) \ dt \]
The solution involves finding the antiderivative of the velocity function and evaluating it from 0 to 5. The result of this calculation gives us the net change in position:
\[ \left[\frac{t^3}{3} - 3t^2 + 8t\right]_0^5 = \frac{10}{3} \text{ m} \]
This value represents how far and in what direction the object has moved from its starting point.
The velocity function given is \( v(t) = t^2 - 6t + 8 \). To find the displacement, we integrate this function:
\[ \int_0^5 (t^2 - 6t + 8) \ dt \]
The solution involves finding the antiderivative of the velocity function and evaluating it from 0 to 5. The result of this calculation gives us the net change in position:
\[ \left[\frac{t^3}{3} - 3t^2 + 8t\right]_0^5 = \frac{10}{3} \text{ m} \]
This value represents how far and in what direction the object has moved from its starting point.
Distance Traveled
Distance traveled differs from displacement because it considers the total path covered, regardless of direction. It is a scalar quantity, meaning it has only magnitude. In cases where the object changes direction, as it does in this scenario, distance traveled can be greater than displacement.
The velocity equation \(v(t) = t^2 - 6t + 8\) changes sign, indicating direction changes. To find the total distance, we integrate the absolute value of the velocity function across each interval of motion.
\[\frac{22}{3} \text{ m}\]
This distance is larger because it includes every part of the object's path, even as it moves backward.
The velocity equation \(v(t) = t^2 - 6t + 8\) changes sign, indicating direction changes. To find the total distance, we integrate the absolute value of the velocity function across each interval of motion.
- For \(0 \leq t < 2\) and \(4 < t \leq 5\), where the velocity is positive, we integrate normally.
- For \(2 < t < 4\), where the motion is negative, the velocity function is reversed.
\[\frac{22}{3} \text{ m}\]
This distance is larger because it includes every part of the object's path, even as it moves backward.
Interval Analysis
Interval analysis involves breaking down the motion into different intervals based on where the velocity changes sign. This helps determine when the object is moving forward or backward.
Solving \( t^2 - 6t + 8 = 0\) gives \(t = 2\) and \(t = 4\), which splits the time into three intervals:
Solving \( t^2 - 6t + 8 = 0\) gives \(t = 2\) and \(t = 4\), which splits the time into three intervals:
- \(0 \leq t < 2\)
- \(2 < t < 4\)
- \(4 < t \leq 5\)
- For \(0 \leq t < 2\), \(v(1) > 0\), indicating positive motion.
- For \(2 < t < 4\), \(v(3) < 0\), indicating negative motion.
- For \(4 < t \leq 5\), \(v(5) > 0\), indicating positive motion again.
Other exercises in this chapter
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