# Lab Report PARAPHRASE ONLY

Anonymous

Question description

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.

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
I WILL ATTACH AVERAGE VELOCITY VS TIME AND AVERAGE TIME VS VELOCITY.

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/s2? 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

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 %

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