Problem

Using approximate values, calculate the slope of …

01:14

Problem 59 Hard Difficulty

(a) By taking the slope of the curve in Figure $2.72,$ verify that the velocity of the jet car is 115 $\mathrm{m} / \mathrm{s}$ at $t=20 \mathrm{s}$ . (b) By taking the slope of the curve at any point in Figure 2.73 , verify that the jet car's acceleration is 5.0 $\mathrm{m} / \mathrm{s}^{2}$ .

Answer

(a) $v=125 \mathrm{m} / \mathrm{s}$
(b) $a=5.0 \mathrm{m} / \mathrm{s}^{2}$

Related Courses

Physics 101 Mechanics

College Physics for AP® Courses

Chapter 2

Kinematics

Discussion

You must be signed in to discuss.

Top Physics 101 Mechanics Educators

Watch More Solved Questions in Chapter 2

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Video Transcript

So this problem is just finding the slope of a line to prove of velocity and acceleration based on the grass. So in the first part, we have a position versus time graph, and it wants us to prove the velocity at 20 seconds. Um, So what we have to do is in order to find the velocity, we have to find this slope. In order to find the slope of the line, we have to pick two points. So let's pick this point. And this point in this case, it does kind of matter what two points you're going to pick because you want to pick two points that are close. Bye to the point. Bigger evaluating. We can't pick the very last point in the very beginning point because this doesn't have a steady slope. So the closer together, these two points are the more accurate or numbers you're going to be. Once we find our two points, we find the change in y over the change in X. So I have that translated into our equation for velocity over here change and why is the same as change in position? So we do our final position minus initial position change in X is the same as changing time, so approximating these numbers at this point, the position is about 2150 meters and at this point it's about 1000 meters. So we find the difference between those two, Um and then we have to take the difference in time. So 25 minus 15 is 10. So when you do this math out, you find that the velocity of 115 meters per second just like it says in the problem. Now, with the next graph, it wants us to show the acceleration. In order to find the acceleration, we again find the slope. But it's the slope of a velocity versus time graph this time not position versus time. So this is a nice, easy one because this is a straight line, meaning it has a constant slope the entire time so we can choose any two points. I happen to choose these two points right here because they're close to the axes and easy to read. But you could choose the first and last or any two random points along this line. Now we have to find her change and wine and divide by change in X, which happens to be the same as a change in velocity, divided by a change in time. So with the two points that I chose a 43 minus 18 divided by the time of five seconds, so we see that the acceleration is five meters.

Get More Help with this Textbook