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$\because$ A sled slides from rest down an icy slope. Measurements takenfrom a video show that the distance from the starting point to thesled at various times is as follows:

(a) Plot the position of the sled versus time. (b) From your plot,determine the approximate average speed of the sled from$t=0.25$ s to $t=1.3 \mathrm{s}$ . (c) Determine the approximate accelera-tion of the sled, assuming it to be constant.

(a) The graph seems to be close to a parabolic curve.(b) 2 $\mathrm{m} / \mathrm{s}$(c) 2.816 $\mathrm{m} / \mathrm{s}^{2}$

Physics 101 Mechanics

Chapter 2

One-Dimensional Kinematics

Motion Along a Straight Line

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Okay, so in this problem, we have a table of measurements and we have to plot the graph off these measurements. This is this is the movement of his lad. So let's see, the first point is, in 0.5 seconds and zero point dirty seeks. Let's see. You know 0.36. It's great to be one. This is one so 0.36 beat something like here. You did it for you. 36. This is the first point. Okay, the second point is one with 1.3. Oh, also is going to be something like here. The next point is 1.5 and 3.2 0.2 is 1233.2, actually, around here, next point is in two seconds and 5.7 I've is 123455.7. Something like you. The next one. And final one is 2.5. Your point. First on, we grows way up into eight points. Eight. So if we connect all the dots and going to have behavior, it is something like a parable. So the first item we plot a graph in we know it sir Credible. The second item we need to tell the average velocity between the times. Judo 0.25 up into time. 1.3. Wrong 0.3 seconds. Okay, so if you look to the graph, 1.3 seconds is actually let's be convicted. Smaller one 0 25 is actually pretty. It's pretty close to zero points. 18. So we can say that the position of 0.25 0 point 18 and the positions of 1.3 just something like you could be close two close to 2.3 2.3 meters so we can calculate the average. It's Pete just going to be 2.3 minors, you know, 0.18 divided by 1.3, minus 0.25. And this is going to be equal to closer to 2.2 meters per second. That's the answer. The first, the second item on the final right. And we need to discover the acceleration well. To discover the acceleration, we just need to use one of the creations of motion. We know that the position as a function off time, it's written us. Excuse us zero t with us a he squared, divided by two. So we know that this light starts from rest. Rich means his initial velocity is going to be zero. So we just need to be one of the segments that we have. Let's pick bigger wrong 2.5 in 8.8. Going to have 8.8 of distance equals a divided by two 2.5 square. So then we have to the acceleration off this light going to be 8.8, divided by 2.5 square times. Two. This is equal to two point 2.8 16 meters burst seconds square, and that's the final answer to this problem.

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