The graph in Fig. E2.31 shows the velocity of a motorcycle police officer plotted as a function

step 1 All right, so I sketched out the graph given in the problem. And then we're in part A, we're fig

The graph in Fig. E2.31 shows the velocity of a motorcycle...

Key Concepts

Instantaneous Acceleration

Instantaneous acceleration is the rate at which velocity changes at a specific moment in time. It is defined mathematically as the derivative of the velocity function with respect to time and provides insight into how quickly an object is speeding up or slowing down at an exact point.

Differentiation

Differentiation is a fundamental concept in calculus used to determine the instantaneous rate of change of a function. In the context of a velocity-time graph, it is used to calculate the instantaneous acceleration by finding the slope of the velocity curve at any given point.

Velocity-Time Graph Analysis

A velocity-time graph visually represents how an object's velocity changes over time. The shape and slope of the graph provide important information: the slope indicates acceleration, and the area under the curve represents the displacement or distance traveled by the object.

Integration

Integration is the mathematical process of accumulating quantities, representing the inverse operation to differentiation. When applied to a velocity function over time, integration calculates the total displacement by summing the infinitesimal changes in position over the given time interval.

Displacement Calculation from Graphs

Displacement, the change in position of an object, can be obtained by computing the area under a velocity-time graph. This involves summing the areas of geometric shapes formed by the graph segments, which is equivalent to integrating the velocity function over a specified time period.

Transcript

00:03 All right, so i sketched out the graph given in the problem.

00:08 And then we're in part a, we're figuring out the accelerations at different points in time.

00:14 And then in part b, we're figuring out the displacement at different points in time.

00:20 So the keys here are that the slope of a vt graph is the acceleration.

00:37 And that can be found by taking the derivative.

00:49 So remember that derivatives of a function get you the slope.

00:55 Now the other thing that we know is that the area under the curve, curve of the vt graph is equal to the displacement.

01:15 And that can be found with spilled.

01:19 That can be found by taking the interval.

01:29 So in this one, because we have constant acceleration, meaning that we have these linear functions.

01:36 We don't have to use integration and derivatives to find these values.

01:42 We can just use geometry and rise over the round of the slope.

01:51 So let's do that.

01:52 So here i have three segments.

01:59 I'm going to call them or actually i have four.

02:01 So i'm just going to call this one a, this segment b.

02:07 Actually, i have three, a, b and c.

02:10 Now, the slope in the first part of our function in the segment a is an acceleration of zero.

02:21 So we can say that since the slope is zero, there's no rise here, a at three seconds is zero.

02:28 Now, the a at seven seconds, that's on the line segment b.

02:33 So we can find the acceleration by taking the rise over run of segment b.

02:41 So rise over run.

02:45 So the rise is from 20 to 45.

02:48 So 45 minus 20, 25.

02:53 Divided by how long does that take? it took from nine seconds from six.

02:59 So three seconds.

03:00 So 25 divided by three will tell us our acceleration.

03:07 Of the police motorcycle.

03:11 So 8 .3 meters per second square.

03:17 I should label that zero as well, zero meters per second squared.

03:22 And then on a lot more on the segment c here.

03:25 So again, it's a rise over run of that.

03:32 So we're talking what, we are going to have a negative 45 here.

03:37 It's going to be 0 minus 45, so negative 45.

03:44 And a run of five.

03:46 So it's going to be negative nine.

03:54 All right, so part b is all about displacement, which we're going to find using geometry.

03:59 We're going to find the area.

04:02 So in five seconds, i'm going to use a different color here.

04:08 I'm going to use blue.

04:11 So if i look at this rectangle right here, this will tell us the first five seconds.

04:22 And then i'm going to use, i'll use blue still, but i'll make it shaded the different direction.

04:32 So we're going to add this part onto the second questions to nine seconds.

04:39 And then the last one is going to be everything but one second worth of time right here.

04:47 So we're just going to find the areas of these different pieces.

04:52 We can use trapezoid rule, we can use rectangles, length times width times, or length times width, or one half case times height or a triangle, all kinds of ways we can do this.

05:08 So it's up to you how you decide which geometry rules you want to follow.

05:13 So i'm just going to do the length times width here for the area.

05:17 So area is equal to length times width.

05:20 And the length here is five, five seconds, and the height or width is going to be 20...

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