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Predict/Calculate An expectant father paces back and forth,producing the position-versus-time graph shown in HGURE $2-32$Without performing a calculation, indicate whether the father'svelocity is positive, negative, or zero on each of the following seg-ments of the graph: (a) $\mathrm{A},$ (b) $\mathrm{B},$ (c) $\mathrm{C}$ , and (d) D. Calculate thenumerical value of the father's velocity for the segments (e) A, (f) B,(g)C, and (h) $\mathrm{D}$ , and show that your results verify your answers toparts $(a)-(d)$

(a) positive(b) zero(c) positive(d) negative(e) 2.0 $\mathrm{m} / \mathrm{s}$(f) 0 $\mathrm{m} / \mathrm{s}$(g) 1.0 $\mathrm{mls}$(h) $-1.5 \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 2

One-Dimensional Kinematics

Motion Along a Straight Line

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well in problems 1 to 1. The 1st 4 items we just need to to answer if the velocity of the father's positive negative or zero for each segment. Well, the only thing we actually needs to know is that if the velocity of the fathers the movement of the father is with constant velocity, then the function that describe his movement is going to be acts of tea was excusable plus fi rusty. Okay, and this is a first degree question or first degree function Better see if and we know that in the first degree function. If V it's bigger than zero, then the function are The function is recent, and if we is smaller than zero, the function is dick recent. So the first item is just to tell the velocity. If he's positive, negative or zero, the first segment there first segment, ask recent so velocity. It's positive. The second segment we don't change. Our position is neither crescent or decreasing, so the velocity zero third segment is also increasing. So the velocity it's bigger than zero and the final segment the velocity. That segment is decreasing, so velocity is smaller than zero. Okay, this is the first part of the problem. The second part of the problem. We want to know the actual value off the velocity of each segment. And how do we calculate the average velocity is just the difference in position, divided by the difference in time. So let's go. The first itin going to be the velocity of the first segment of the last of the first segment is just difference in position. So we go from position one positions. Sorry, we go from position zero shown position too. And we spent one second during this. So the velocity of a is just too meters per second. The velocity off the second segment is zero because we don't change your position and any moment, the velocity off the third position. We just go from position to from position tree in also one second. So our average velocity is one meters per second and the velocity of the final segment. It's just going to be Let's see a D in here. Now it's H. Let's see, we go from position tree to position zero. So we go zero minus tree in How much seconds in two seconds. Two seconds. So we have a negative velocity off 1.5 meters per second. This is the final answer. Thank you for watching

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