Problem 8
Question
Suppose that the velocity of a moving object is given by \(v(t)=t(t-1)(t-3),\) measured in feet per second, and that this function is valid for \(0 \leq t \leq 4\) a. Write an expression involving definite integrals whose value is the total change in position of the object on the interval [0,4] . b. Use appropriate technology (such as http://gvsu.edu/s/a93) to compute Riemann sums to estimate the object's total change in position on [0,4] . Work to ensure that your estimate is accurate to two decimal places, and explain how you know this to be the case. c. Write an expression involving definite integrals whose value is the total distance traveled by the object on [0,4] d. Use appropriate technology to compute Riemann sums to estimate the object's total distance travelled on [0,4] . Work to ensure that your estimate is accurate to two decimal places, and explain how you know this to be the case. e. What is the object's average velocity on [0,4] , accurate to two decimal places?
Step-by-Step Solution
Verified Answer
1. Integrate \(v(t)\) for total change in position. 2. Use Riemann sums to estimate integrals for total change in position and total distance traveled. 3. Calculate average velocity using the integral of \(v(t)\).
1Step 1: Recognize the velocity function
The velocity of the moving object is given by the function \( v(t) = t(t-1)(t-3) \) measured in feet per second within the interval \(0 \leq t \leq 4\).
2Step 1: Total Change in Position
To find the total change in position, we need to evaluate the definite integral of the velocity function over the interval [0, 4]. The expression is: \[ \text{Total Change in Position} = \ \ \ \ \ \ \ \ \int_{0}^{4} v(t) \, dt = \ \ = \ \ \ \ \ \ \ \int_{0}^{4} t(t-1)(t-3) \, dt \ \]
3Step 2: Estimate Total Change in Position with Riemann Sums
Use technology to compute Riemann sums. Here is an example of how you can do it using a midpoint Riemann sum with \(n=100\) subintervals: Divide the interval [0,4] into 100 equal subintervals - Calculate the sum using the expression \( \sum_{i=1}^{100} v(t_i)\Delta t \), where \( t_i \) is the midpoint of the subinterval and \( \Delta t = \frac{4-0}{100}\). Ensure that the estimate is accurate to two decimal places.
4Step 3: Total Distance Traveled
To find the total distance traveled, integrate the absolute value of the velocity function over the interval [0, 4]. The expression is: \[ \text{Total Distance Traveled} = \ \ \ \ \ \ \ \int_{0}^{4} |v(t)| \, dt = \ \ = \ \ \ \ \ \ \ \int_{0}^{4} |t(t-1)(t-3)| \, dt \ \]
5Step 4: Estimate Total Distance Traveled with Riemann Sums
Use technology to compute Riemann sums. Here is an example of how you can do it using a midpoint Riemann sum with \(n=100\) subintervals: Divide the interval [0,4] into 100 equal subintervals - Calculate the sum using the expression \( \sum_{i=1}^{100} |v(t_i)|\Delta t \), where \( t_i \) is the midpoint of the subinterval and \( \Delta t = \frac{4-0}{100}\). Ensure that the estimate is accurate to two decimal places.
6Step 5: Average Velocity
To find the average velocity of the object over the interval [0, 4], use the formula: \[ \text{Average Velocity} = \frac{1}{b-a} \int_{a}^{b} v(t) \, dt = \frac{1}{4-0} \int_{0}^{4} t(t-1)(t-3) \, dt \ \]
Key Concepts
definite integralsvelocity functionRiemann sumstotal distanceaverage velocity
definite integrals
A definite integral is a mathematical tool that accumulates total change over a specific interval. When evaluating a definite integral, you're essentially summing up small areas under a curve from one point to another. This can be used to find things like the total change in an object's position by integrating its velocity function over time.
For example, to find the total change in position of an object with the velocity function \(v(t) = t(t-1)(t-3)\) between \(t=0\) and \(t=4\), we use the definite integral: \[ \text{Total Change in Position} = \int_{0}^{4} v(t) \, dt = \int_{0}^{4} t(t-1)(t-3) \, dt \] This integral helps us calculate how much the object's position has changed over that 4-second interval.
For example, to find the total change in position of an object with the velocity function \(v(t) = t(t-1)(t-3)\) between \(t=0\) and \(t=4\), we use the definite integral: \[ \text{Total Change in Position} = \int_{0}^{4} v(t) \, dt = \int_{0}^{4} t(t-1)(t-3) \, dt \] This integral helps us calculate how much the object's position has changed over that 4-second interval.
velocity function
The velocity function describes how an object's speed and direction change over time. In our example, the velocity function \(v(t) = t(t-1)(t-3)\) tells us how fast and in what direction the object is moving at any time \(t\).
Understanding the velocity function is crucial because it helps us integrate and find other important values like total distance and average velocity. By knowing how the object's speed changes, we can calculate positional changes using tools like definite integrals.
Understanding the velocity function is crucial because it helps us integrate and find other important values like total distance and average velocity. By knowing how the object's speed changes, we can calculate positional changes using tools like definite integrals.
Riemann sums
Riemann sums are a way to approximate the value of a definite integral. We divide the interval into smaller sub-intervals, calculate the function's value at specific points within these sub-intervals, and then sum these values to get an approximation.
For instance, to estimate the total change in position using a midpoint Riemann sum, we divide the interval [0,4] into 100 sub-intervals. The width of each sub-interval is \( \Delta t = \frac{4-0}{100} = 0.04\ \). The midpoint for each sub-interval is calculated, and we sum up the function values at these midpoints: \[ \sum_{i=1}^{100} v(t_i) \Delta t \] This method gives an estimate that can be quite close to the true value, especially with a larger number of sub-intervals.
For instance, to estimate the total change in position using a midpoint Riemann sum, we divide the interval [0,4] into 100 sub-intervals. The width of each sub-interval is \( \Delta t = \frac{4-0}{100} = 0.04\ \). The midpoint for each sub-interval is calculated, and we sum up the function values at these midpoints: \[ \sum_{i=1}^{100} v(t_i) \Delta t \] This method gives an estimate that can be quite close to the true value, especially with a larger number of sub-intervals.
total distance
The total distance traveled by an object can be different from the total change in position because it considers all movement, regardless of direction. To compute the total distance, we integrate the absolute value of the velocity function over the given interval.
For our velocity function \(v(t) = t(t-1)(t-3)\) over the interval [0, 4], the total distance is given by: \[ \text{Total Distance Traveled} = \int_{0}^{4} |v(t)| \, dt = \int_{0}^{4} |t(t-1)(t-3)| \, dt \] By taking the absolute value, we ensure that all distances are positive, summing up all motion without canceling out movements in opposite directions.
For our velocity function \(v(t) = t(t-1)(t-3)\) over the interval [0, 4], the total distance is given by: \[ \text{Total Distance Traveled} = \int_{0}^{4} |v(t)| \, dt = \int_{0}^{4} |t(t-1)(t-3)| \, dt \] By taking the absolute value, we ensure that all distances are positive, summing up all motion without canceling out movements in opposite directions.
average velocity
The average velocity gives an overall picture of how fast an object is moving over a specific interval. It is calculated by taking the total change in position and dividing it by the total time interval.
For our example, where \(v(t) = t(t-1)(t-3)\) over the interval [0, 4], the average velocity is found using: \[ \text{Average Velocity} = \frac{1}{b-a} \int_{a}^{b} v(t) \, dt = \frac{1}{4-0} \int_{0}^{4} t(t-1)(t-3) \, dt \] This value gives a simple measure of the object's typical speed over the interval, providing insight into its performance and efficiency.
For our example, where \(v(t) = t(t-1)(t-3)\) over the interval [0, 4], the average velocity is found using: \[ \text{Average Velocity} = \frac{1}{b-a} \int_{a}^{b} v(t) \, dt = \frac{1}{4-0} \int_{0}^{4} t(t-1)(t-3) \, dt \] This value gives a simple measure of the object's typical speed over the interval, providing insight into its performance and efficiency.
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