Approximating displacement The velocity in ft/s of an obj...

Question

Approximating displacement The velocity in ft/s of an object moving along a line is given by v = 3t² + 1 on the interval 0 ≤ t ≤ 4, where t is measured in seconds.

(a) Divide the interval [0,4] into n = 4 subintervals, [0,1] , [1.2] , [2,3] , and [3,4]. On each subinterval, assume the object moves at a constant velocity equal to v evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on [0, 4] (see part (a) of the figure)

Accepted Answer

Hello. In this video, we are told that a particle moves along a straight line with velocity in meters per second. It is given by the function V of T is equal to 222 + 4 fort in 1 to 5. We want to divide the interval from 1 to 5 into 4 equal subintervals, and on each subinterval, assume that the particle's velocity remains constant at the value of VFT evaluated at the midpoint of the sub-intervals. Using these constant velocity subintervals, approximate the displacement of the particle from 1 to 5. So, let's just go ahead and list out the given information. First, we are given the velocity function, which is defined as V of T is equal to 2 T 2 + 4. And we are working on the interval from 1 to 5. Furthermore, we are going to split this interval into 4 equal subintervals. So how can we solve for the approximate displacement of the particles? Well, what we can do is we can go ahead and sum up the velocities of the midpoints. So here, the first thing we want to do is we want to go ahead and define the width of each of the subintervals. That is going to be defined as delta T, and delta T is defined as the larger boundary of our interval minus the smaller boundary divided by the amount of subintervals we are working with. That is going to be 5 minus 1 divided by 4, which simplifies to 4 divided by 4, which further simplifies to 1. And so what this means is that the width of each of the sub-intervals is going to be one. Now let's go ahead and define the midpoints at each of the sub-intervals. Now, the sub-intervals are the points between each interval from 1 to 22 to 33 to 4, and 4 to 5. Those values are going to be 1.5, 2.5, 3.5, and 4.5. By using the graph, the value at 1.5 is going to be approximately 8.5 m. The value at 2.5 is going to be approximately 16.5. The value at 3 or 3.5 is going to be approximately 28.5, and the value at 4.5 is going to be approximately 44.5. And so what we want to do now is you want to go ahead and just write a summation for each of the velocities in the sub-interval. That summation is going to be the summation starting at 1 and going to 4, a V of X sub I multiplied by delta T. This is going to further expand to be Delta T multiplied by V of 1.5. Plus V of 2.5. Plus V of 3.5. Plus V of 4.5. And so by plugging in all the values that we've solved for, we can rewrite the summation to be one, multiplied by each of the velocities at the midpoint, which is going to be 8.5 plus 16.5, plus 28.5, plus 44.5. And with the help of a calculator, this is going to give us a total velocity, or total displacement of 98 m. So this is going to be the total displacement with the given graph, by using the Ryman sums of the midpoints. And with that being said, the overall solution to this problem is going to be B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Key Concepts

Velocity Function

Midpoint Rule

Displacement

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