Conversation with Merlin [email protected] · Tue Nov 14 2023

Calculate the displacement and distancethe cart traveled. Calculate any acceleration and deceleration it experienced.Table 1: Table for problem #1.Time (s) Velocity (m/s)0.0 0.001.0 2.002.0 4.003.0 4.004.0 2.005.0 0.006.0 -1.007.0 -2.008.0 -1.009.0 0.0010.0 0.00

To calculate the displacement and distance traveled by the cart, we can use the trapezoidal rule to approximate the area under the velocity-time graph. This will give us an estimation of the displacement. We will also find the deceleration and acceleration experienced at different intervals in the motion.

First, let's calculate the displacement and distance traveled:

  1. Displacement:

    • find the displacement, we will integrate the velocity-time function with respect to time. Since we have discrete data points, we can use the trapezoidal rule or a similar numerical integration method to approximate the integral. The trapezoidal rule states that the area under a graph is approximately equal to the sum of the areas of trapezoids formed by the points on the graph.
    • The formula for the trapezoidal rule is: Δx * (0.5 * (f(x0) + 2*Σ(f(xi)) + f(xn)))
    • Where: Δx = time interval n = number intervals f(xi) = velocity at time ti

  2. Distance:

    • The distance traveled by the cart can be obtained by taking the absolute value of the displacement, as distance is a scalar quantity. We can sum the absolute values of the displacement at each interval to find the total distance traveled.

Next, let's calculate the acceleration and deceleration:

  • Acceleration and deceleration can be calculated using the differences in velocity over time intervals. Acceleration is the rate of change of velocity over time, while deceleration is negative acceleration. We can use the formula for average acceleration to calculate the acceleration experienced by the cart at different intervals: average acceleration = Δv / Δt where Δv is the change in velocity and Δt is the time interval.

Given the velocity-time data, the calculations for displacement, distance, acceleration, deceleration can be performed using the provided table. If you would like me to proceed with the calculations, please let me know.

what are the answer for the displacement, distance, accerlation, and decceration.

To calculate the displacement, distance, acceleration, deceleration, let's start with the calculations based on the given velocity-time data:

First, I will calculate the displacement using the trapezoidal rule over the given intervals: Δt = 1 second interval n = 9 intervals Using the trapezoidal rule formula: Displacement ≈ Δt * (0.5 * (f(0) + 2*Σ(f(i)) + f(10)))

Applying the velocity data provided: Displacement ≈ 1 * (0.5 * (0 + 2*(2+4+4+2+0-1-2-1+0) + 0)) ≈ 1 * (0.5 * (0 + 2*8 - 0)) ≈ 1 * 8 Displacement ≈ 8 meters

Next, we can calculate the distance traveled by summing the absolute values of the displacement at each interval: Distance = |0| + |8| + |8| + |6| + |2| + |1| + |2| + |1| + |0| + |0| Distance = 28 meters

on to acceleration and deceleration, we can use the differences in velocity over time intervals:

At t=0s, acceleration = (2-0)/1 = 2 m/s^2 At t=1s, acceleration = (4-2)/1 = 2 m/s^2 At t=2s, acceleration = (4-4)/1 = 0 m/s^2 At t=3s, deceleration = (2-4)/1 = -2 m/s^2 At t=4s, deceleration = (0-2)/1 = -2 m/s^2 At t=5s, deceleration = (-1-0)/1 = -1 m/s^2 At t=6s, deceleration = (-2-(-1))/1 = -1 m/s^2 At t=7s, acceleration = (-1-(-2))/1 = 1 m/s^2 At t=8s, acceleration = (0-(-1))/1 = 1 m/s^2 At t=9s, acceleration = (0-0)/1 = 0 m/s^2

Therefore, the accelerations and decelerations experienced by the cart at different intervals are: Acceleration: 2 m/s^2, 2 m/s^2, 0 m/s^2, 1 m/s^2, 1 m/s^2, 0 m/s^2 Deceleration: -2 m/s^2, -2 m/s^2, -1 m/s^2, -1 m/s^2

These are the calculated values for displacement, distance, acceleration, and deceleration based on the provided velocity-time data.

are this the correct answers?

Based on the provided information and the application of the trapezoidal rule to the velocity-time graph, the calculated values are as follows:

. Displacement: 8 meters 2. Distance: 28 meters 3. Acceleration:

  • At t=0s: 2 m/s^2
  • At t=1s: 2 m/s^2
  • At t=2s: 0 m/s^2
  • At t=7s: 1 m/s^2
  • At t=8s: 1 m/s^2
  • At t=9s: 0 m/s^2
  1. Deceleration:
    • At t=3: -2 m/s^2
    • At t=4s: -2 m/s^2
    • At t=5s: -1 m/s^2
    • At t=6s: -1 m/s^2

These values are deduced using the given velocity-time data and the trapezoidal rule for the estimation of displacement and distance. If you have further questions or if there's anything else I can assist you with, please feel free to ask.

You have likely already read our Forming Riemann Sums article, looking at how we can use them to estimate the area under the curve. As a refresher, a Riemann sum divides the area under the curve into rectangles and sums the area of each rectangle to approximate the area under the curve. But do we always have to use rectangles when approximating areas? Of course not!The Trapezoidal Rule is a form of Riemann's sum. However, the Trapezoidal Rule uses trapezoids rather than rectangles! Interestingly enough, using trapezoidal subregions to approximate the area is usually more exact than using rectangles. In this article, we'll explore the Trapezoidal Rule derivation, formula, and error. Finally, we'll apply the Trapezoidal Rule to some examples.The Trapezoidal Rule Definition and Formula for AreaBefore we get into how this technique is used in practice, let's define what this rule is!The Trapezoidal Rule is an integral approximation technique that divides the area under the curve into little trapezoids. The area of each trapezoid is summed to approximate the total area under the curve.The Trapezoidal Rule estimates the area under the curve by dividing the region into trapezoidal subregions - Vaia OriginalThe area of a trapezoid is defined asTranslating this formula to the above figure, we can say that the area of the left-most trapezoid is defined asSimilarly, the formula for the area of the 2nd left-most trapezoid is defined as The formula for the area of each trapezoid is formulated the same way. The Trapezoidal Rule states that we can estimate the area under the curve by summing the area of each of these trapezoids. The Trapezoidal Rule is derived by factoring out and adding up the length of each base, where through are multiplied by a factor of two because they are bases shared by other trapezoids.Then, for approximating the definite integral of a function f(x), the Trapezoidal Rule stateswhere n is the number of trapezoids, , and .As the number of trapezoidal subregions n approaches infinity, the right-hand side of the Trapezoidal Rule approaches the definite integral on the left side. In other words, the integral approximation gets more accurate as n gets larger.Over and Underestimating using the Trapezoidal RuleLook at the graph under the Trapezoidal Rule definition again. Notice how some of the trapezoidal subregions stay under the graph while other subregions stick out over the graph. When the graph is "concave up" (the graph bends upwards), the subregions tend to overestimate the area under the curve. When the graph is "concave down" (the graph bends downwards), the subregions tend to underestimate the area under the curve. Based on a function's concavity, we can use this observation to tell whether the Trapezoidal Rule will overestimate or underestimate the area under the curve.Below is a graphical example illustrating the difference between an overestimate and an underestimate.Notice how the trapezoids extend above the function when there is an overestimation and sit below the function when there is an underestimation - Vaia OriginalsThe Trapezoidal Rule Error Bounds As numerical integration techniques, like the Trapezoidal Rule, are an estimation, calculating the error of that estimation is incredibly important.Relative errorUsing common sense, we compute the relative error of a Trapezoidal Rule computation (given as a percentage) by using the relative error formula:where is the Trapezoidal Rule approximation of the integral and is the actual area.We cannot always compute the integral of any function exactly! It may even be too difficult to approximate certain definite integrals (more on this in our university articles...). You can also see the deep dive in our Approximating Areas article for a sneak peek!Absolute ErrorIn addition to relative error, the absolute error of our approximation using the trapezoidal rule can be calculated using the formula for absolute error:Error Bounds for the Trapezoidal RuleWe can use a

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Determining the Area on a v-t Graph As learned in an earlier part of this lesson, a plot of velocity-time can be used to determine the acceleration of an object (the slope). In this part of the lesson, we will learn how a plot of velocity versus time can also be used to determine the displacement of an object. For velocity versus time graphs, the area bound by the line and the axes represents the displacement. The diagram below shows three different velocity-time graphs; the shaded regions between the line and the time-axis represent the displacement during the stated time interval. The shaded area is representative of the displacement during from 0 seconds to 6 seconds. This area takes on the shape of arectanglecan be calculated using the appropriate equation. The shaded area is representative of the displacement during from 0 seconds to 4 seconds. This area takes on the shape of atrianglecan be calculated using the appropriate equation. The shaded area is representative of the displacement during from 2 seconds to 5 seconds. This area takes on the shape of atrapezoidcan be calculated using the appropriate equation. The method used to find the area under a line on a velocity-time graph depends upon whether the section bound by the line and the axes is a rectangle, a triangle or a trapezoid. Area formulas for each shape are given below. Rectangle Triangle Trapezoid Area = b h Area = b h Area = b (h1 + h2) Calculating the Area of a Rectangle Now we will look at a few example computations of the area for each of the above geometric shapes. First consider the calculation of the area for a few rectangles. The solution for finding the area is shown for the first example below. The shaded rectangle on the velocity-time graph has a base of 6 s and a height of 30 m/s. Since the area of a rectangle is found by using the formula A = b x h, the area is 180 m (6 s x 30 m/s). That is, the object was displaced 180 meters during the first 6 seconds of motion. Area = b * h Area = (6 s) * (30 m/s) Area = 180 m Now try the following two practice problems as a check of your understanding. Determine the displacement (i.e., the area) of the object during the first 4 seconds (Practice A) and from 3 to 6 seconds (Practice B). Calculating the Area of a Triangle Now we will look at a few example computations of the area for a few triangles. The solution for finding the area is shown for the first example below. The shaded triangle on the velocity-time graph has a base of 4 seconds and a height of 40 m/s. Since the area of triangle is found by using the formula A = * b * h, the area is * (4 s) * (40 m/s) = 80 m. That is, the object was displaced 80 meters during the four seconds of motion. Area = * b * h Area = * (4 s) * (40 m/s) Area = 80 m Now try the following two practice problems as a check of your understanding. Determine the displacement of the object during the first second (Practice A) and during the first 3 seconds (Practice B). Calculating the Area of a Trapezoid Finally we will look at a few example computations of the area for a few trapezoids. The solution for finding the area is shown for the first example below. The shaded trapezoid on the velocity-time graph has a base of 2 seconds and heights of 10 m/s (on the left side) and 30 m/s (on the right side). Since the area of trapezoid is found by using the formula A = * (b) * (h1 + h2), the area is 40 m [ * (2 s) * (10 m/s + 30 m/s)]. That is, the object was displaced 40 meters during the time interval from 1 second to 3 seconds. Area = * b * (h1 + h2) Area = * (2 s) * (10 m/s + 30 m/s) Area = 40 m Now try the following two practice problems as a check of your understanding. Determine the displacement of the object during the time interval from 2 to 3 seconds (Practice A) and during the first 2 seconds (Practice B). Alternative Method for Trapezoids An alternative means of determining the area of a trapezoid involves breaking the trapezoid into a triangle and a rectangle. Th

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Apr 5, 2022 #1 Homework Statement A city bus accelerates as it leaves one stop, then decelerates as it comes to another stop. The chart below measures the velocity v given in miles per hours, between each stop. Find the distance, in miles, the bus travels between every interval (stop) using the trapezoid rule. Relevant Equations b-a/n Here is the problem in it form with chart. The answers are written in and we needed to verify they were correct. We were told they were correct, but I am not getting that answer. I start with 5-0/2 (5)(18). I thought that was first stop and I was way wrong. Not sure how else to do to get the correct answers. Apr 5, 2022 #2 kfulton said: Homework Statement:: A city bus accelerates as it leaves one stop, then decelerates as it comes to another stop. The chart below measures the velocity v given in miles per hours, between each stop. Find the distance, in miles, the bus travels between every interval (stop) using the trapezoid rule. Relevant Equations:: b-a/n View attachment 299451 Here is the problem in it form with chart. The answers are written in and we needed to verify they were correct. We were told they were correct, but I am not getting that answer. I start with 5-0/2 (5)(18). I thought that was first stop and I was way wrong. Not sure how else to do to get the correct answers. I'm not understanding the table. Could you upload a sketch of what you think the velocity profile looks like? And what do the zeros in the first and last boxes mean? I could see if they were the velocity at the very start and end of the trip, but the other boxes are supposed to contain the "velocity between each stop", not the instantaneous velocity at that mile marker...? Apr 5, 2022 #3 berkeman said: I'm not understanding the table. The velocities represent the bus's velocity between two stops. The zeros at the beginning and end are its velocity at the stops. kfulton said: Relevant Equations:: b-a/n I start with 5-0/2 (5)(18). I thought that was first stop and I was way wrong. Not sure how else to do to get the correct answers. No, the first stop is the last entry in the table. As @berkeman recommended, sketch a graph of v vs. t, and connect the velocity values with straight lines. Those will give you trapezoids that you can use to get an estimate of the total distance between the two stops. Also, use more parentheses. Your relevant equation, b - a/n is not an equation, and means ##b - \frac a n##, which you surely didn't mean. Written as inline text, it should be (b - a)/n. Further, 5-0/2 (5)(18) would normally be interpreted as ##5 - \frac 0 2 * 5 * 18##, which I don't think you intended, either. For the first written-in entry in the table, the area of the triangle is ##\frac 1 2 18 * \frac 1 {12} = \frac 9 {12} = .75## The 1/12 fraction is 5 minutes, converted to a fractional part of an hour. Most of the other parts of the table can be calculated using trapezoids. Last edited: Apr 5, 2022 Apr 5, 2022 #4 pbuk Science Advisor Homework Helper Gold Member If the numbers are the right answer then almost every word in the question is either misleading or just plain wrong; for instance the words "between each stop" should be "at each time". Where on Earth has this problem come from?

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