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Answer analysis part 1 and part 2 show work & clear enough to read!
PHY 1100- Exercise 2- Report Name:| Part 1 -Velocity Position versus Time (x vs t0 graph for Object 1 x versus t- Object 1 y 1.2754x+0.0187 Linear x) t (s) Position versus Time (x vs t) graph for Object 2 x versus t- Object 2 4.5 3.5 EEETE0.7249%+ 0.0513 2.5 Linear (x) 1.5 0.5 t (s)
Position versus Time (x vs t) graph for Both Objects x versus t for Both Objects y 0.7652x+ 0.0187 Linear (x) Linear x) -0.4349x+0.0513 0.6 12 18 24 642 4 54 t (s) Part 2 Acceleration Position versus Time (x vs t) graph for Object 3 x versus t- Object 3 45 40 35 30 25 食20 1.7909x 9.272 10 Linear (x) 2025 -10 t (s)
Position versus Time squared (x vs th) graph for Object 3 x vs tA2 700 600 500 400 y-15.887x-4.6416 200 Linear (t) 100 10 20 30 40 50 t (s)
Analysis Part 1 1) What are the slopes for Object 1 and Object 27 2) What does the fact that the plots are linear tell us about the velocity of these objects? 3) Which of the two objects was traveling faster? Explain your reasoning. Part 2 1) Which of the two graphs is linear? What does the fact that this particular graph is linear tell us about the acceleration of the object? (Hint: the position of an object undergoing constant acceleration from rest can be described by the following function:a 2) 3) What was the acceleration of Object 3?
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Answer #1

1) slope of object 1 = 1.2754 ( from graph)

slope of object 2 = .7249 ( from graph )

2) Linear graph represents that there is no acceleration in object.

3) Object 1 moves faster because slope is higher that is velocity.

Part - 2

1) x vs t^2 graph is linear

2) it means there is uniform acceleration in object 3.

3) acceleration of object 3 was 31.774 unit. 2*slope = acceleration  

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