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A bus makes a trip according to the position-time…

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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 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88

Problem 68 Hard Difficulty

A bus makes a trip according to the position-time graph shown in the drawing. What is the average velocity (magnitude and direction) of the bus during each of the segments $A, B,$ and $C$ ? Express your answers in $\mathrm{km} / \mathrm{h}$ .

Answer

$-2.0 \times 10^{1} k m / h$
$-1.0 \times 10^{1} k m / h$
40$k m / h$

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Physics 101 Mechanics

Physics

Chapter 2

Kinematics in One Dimension

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Alyssa J.

October 2, 2021

Part B is +10 km/h

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Video Transcript

this problem is asking us to find the velocity, um, of a bus during its trip based on a position versus time graph. Now the key thing here is that velocity from a position versus time graph is equal to the slope slope of the position versus time graph. So that means for all three sections A, B and C. We're just looking at the slope of the line now. One thing to note here is that that means right away you can tell if these air going to be positives or negatives, because if the line is going downwards, it's a negative slope. So negative velocity. If the line is going upwards, it's a positive slope, so a positive velocity. So my final answers for all of these, if we're looking just at the sign, is a will be a negative velocity. But BNC will both be positives. Now. My rise overrun for all these forgetting the slopes just means we need to find a part of the graph where you can easily measure a change in velocity and a change in time. Um, if we look at the first graph, the first one second sees a change of velocity of 20 meters, so that one is going to be our rise of negative 20 over our run of one second. So we get a velocity of negative 20 meters per second for looking at B. It takes one second for it to go from 10 to 20. So it has a positive change of 10 in one second. So it's traveling at 10 meters per second and for part C, we don't have a nice one. Second interval. We can look at here. The closest we can get is 2/2 of a second and in half a second that see actually goes up by 20. So 20 divided by 0.5 gives us a positive 40 meters per second for our velocity.

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John D. Cutnell, Kenneth W. Johnson

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