Question

HELP NEEDED!!!!!!! especially with the graphing for objects 1 and 2 together and object 3 squared. People are getting different graphs and I do not know if it is correct. Also SIG FIGS in analysis. THANK YOU IN ADVANCE

Part 1 - Velocity Position versus Time (x vs t) graph for Object 1 x versust-Object 1 y 1.2754x+ 0.0187 x (m) Linear (x (m)) 0 Position versus Time (x vs t) graph for Object 2 x versus t-Object 2 4.5 y 0.7249x+ 0.0513 3.5 E 2.5 Linear (x (m) 1.5 0.5 Position versus Time (x vs t) graph for Both Objects x versust-Objects 1 and 2 10 1.2754x+0.018 0.7249x+0 x2 (m)." Linear (xl (m) Linear(2 (m) xd (m) Part 2-Acceleration Position versus Time (x vs t) graph for Object 3 x versus t-Object 3 45 35 E 25 x 20 15 10 x (m) 10 5 15 20 25 30 Position versus Time squared (x vs t2) graph for Object 3 x versus t2 -Object 3 45 y:0.0628x +0.3243 40 35 30 25 x 20 15 10 x(m) Linear ((m)) 0 t2 (s2) Analysis Part 1 1) What are the slopes for Object 1 and Object 2? The slope for object one is 1.2754 m/s. The slope for object two is 0.7249 m/s 2) What does the fact that the plots are linear tell us about the velocity of these objects? The fact that the plots are linear shows us that the velocity is constant. In the case of this graph the average velocity is the slope and it is positive and constant. 3) Which of the two objects was traveling faster? Explain your reasoning. Object 1 was traveling faster because it had a larger change in slope. The slope is change in position (x) over time. This shows that the numerator, which is x, is greater and therefore the object travels faster. You can also tell by looking at the graph and seeing that it is steeper. Part 2 1) Which of the two graphs is linear? The graph of position versus time squared was linear 2) 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: x- iat'] This means that there is uniform acceleration for Object 3. 3) What was the acceleration of Object 3? The acceleration is equal to 2 times the slope. From class we know that x-(at*2)/2. Solving for a you would get 2(x-t^2). X is the y-axis and tA2 is the x-axis; rise over run is the slope. Multiplying the slope by two I would get 0.1258 m/s 2.

Answer 1

Position versus Time (x vs t) graph for Object 1 (ooven xversust-Object 1 1.27 592+0.0187 1.2754x + 0.0187 near ((mil t(s) Position versus Time (x vs t) graph for Object 2 x versust -Object2 4.5 0.7249x+0.0513 slope 724 3.5

Part 1 1) What are the slopes for Object 1 and Object 2? The slope for object one is 1.2754 m/s. The slope for object two is 0.7249 m/s. 2) What does the fact that the plots are linear tell us about the velocity of these objects? -> As c The fact that the plots are linear shows us that the velocity is constant. In the case of this graph the average velocity is the slope and it is positive and constant Post Hve 3) Which of the two objects was traveling faster? Explain your reasoning. Object 1 was traveling faster because it had a larger change in slope. The slope is change in position (x) over time. This shows that the numerator, which is x, is greater and therefore the object travels faster. You can also tell by looking at the graph and seeing that it is steeper

Part 2 - Acceleration Position versus Time (x vs t) graph for Object 3 x versus t-Object 3 Position versus Time squared (xvs t) graph for Object 3 x versust2-0bject 3 45 40 35 Y 0.062803243 25 x 20 Linear (m

xversust2-Object 3 45 40 35 30 25 20 15 10 x . 141 다 '3243 y 0.0628x + 0.3243 near ((m)) t? (s2)