Lab Report PARAPHRASE ONLY
1. How does displacement change with time for a falling object? How might you describe the mathematical relationship between the distance and time of a falling object? Explain the answer using your data, graphs, and kinematic equations.
Answer:The displacement increases as time increases. Even though displacement is increasing with at a constant rate time isn't. For free falling objects with constant acceleration of 9.8 the time is squared with each increasing distance., Using the kinematics equation this can be expressed as d = Vi* t + ½ g *t^2 . Since the ball is starting from rest, the initial time, velocity and displacement are zero. So the formula can be changed to d= 1/2gt^2. The graph of this both the theoretical and real results are quadratic and have a curved trend line. We can also use another kinematics formula if we don't know the acceleration and only have the average velocities, ∆d=1/2(vf+vi)∆t . This formula gives us a linear graph with both time and displacement increasing together.
Theoretical blue and real experiment black. For the
equationd = Vi* t
+ ½ g *t^2
I WILL ATTACH TIME VS DISPLACEMENT GRAPH 1 AND 2
Theoretical and real in blue and black
color respectively for the equation ∆d=1/2(vf+vi)∆t
I WILL ATTACH DISPLACMENT VS TIME
Displacement |
Time |
Time ^2 |
0.10 |
0.43 |
0.18 |
0.20 |
0.65 |
0.38 |
0.30 |
1.04 |
1.08 |
0.40 |
1.23 |
1.51 |
0.50 |
1.64 |
2.68 |
0.60 |
1.87 |
3.49 |
0.70 |
2.18 |
4.75 |
0.80 |
2.44 |
5.95 |
0.90 |
2.54 |
6.45 |
1.0 |
2.87 |
8.24 |
1. How does the velocity of a falling object change with time? How would you describe the mathematical relationship between velocity and time of a falling object? Explain the answer using your data, graphs, and the kinematic equations.
Answer:
Mostly velocity increases as time increases but there were instances where velocity would decrease in the middle. One the graph is plotted the trend line is a straight line so we can say they have a close to linear relationship. Theoretically velocity of of free falling objects constantly increases with time and therefore has a constant acceleration. The kinematics formula that relates velocity and time is Vf=Vi+g∆t[img width="2" height="2" src="file:///C:/Users/Spare/AppData/Local/Temp/msohtmlclip1/01/clip_image003.png" v:shapes="_x0000_i1025">g∆t
Time |
Average- velocity |
acceleration |
Theoretical Average-velocity |
0.43 |
0.23 |
0.53 |
4.21 |
0.65 |
0.31 |
0.48 |
6.37 |
1.04 |
0.29 |
0.28 |
10.19 |
1.23 |
0.33 |
0.27 |
12.54 |
1.64 |
0.3 |
0.18 |
16.07 |
1.87 |
0.32 |
0.17 |
18.32 |
2.18 |
0.32 |
0.15 |
21.36 |
2.44 |
0.33 |
0.14 |
23.91 |
2.54 |
0.35 |
0.14 |
24.89 |
2.87 |
0.35 |
0.12 |
28.12 |
1. Compare the slope of the velocity-time graph to the average of all your acceleration values. Are they close? What does the slope of a velocity (or speed) vs. time graph mean? Explain the answer using your data. How does the value of g that you calculated compare to the accepted value of 9.80 m/s^{2}? What is your percent error? Remember that the value of g can be calculated by using g = a/sinθ and percent error can be calculated using the following equation:
%error = ( I measured value - accepted value I / accepted value ) x 100
Answer:
The slope of my actual lab result graph matches the average of the acceleration of the values listed in the table above which was 0.33 m/s^2. That is because the slope of a velocity-time graph is equal to the average acceleration. The slope of the theoretical velocity-time graph was 9.8 m/s^2.
The g I calculated was 10.1, compared to the accepted value of g it is a bit greater. The percent error is 3.06 %