Lab 8: Motion in One Dimension with Air Drag

Purpose: To analyze how changing force affects motion in one dimension.

Introduction: 
So far we have assumed objects move with constant acceleration. This assumes that a constant force is applied to the object. What if a changing force (which causes a changing acceleration) is applied to an object?

An object under the influence of air drag is one example of this. Calculus provides a way to solve this problem where variables are changing with respect time. A spreadsheet can be used to perform this calculus. Two basic relationships can be used to analyze the object:  

                                                           v_new = v_old + a_avg△t     (1)


Where v_old  is the previous value of velocity and position, and a_avg  is the average value of the acceleration and velocity during the time interval △t.


1. By unit analysis, show that the above equation (1) is valid:

            Left side unit = m/s
            Right side unit = (m/s) + (m/s^2) × s = (m/s) + (m/s) = m/s
          
    So the units of two sides are identical.
 

2. Why do we use a_avg in equation (1).
    
            Because the can be changing all the time, so the average data is more accurate.


3. Come up with an analogous equation relating y_new and y_old:

            y_old = y_o
             
                y_new = y_o + v_old△t + (1/2)a_avg(△t)^2
            or   y_new = y_o + v_avg△t


4. What is the benefit of changing a small △t:

            Because the acceleration is changing all the time, so in a small △t, the a_avg is closer to real value.     So, the v_new will also be more accurate.



Since the acceleration depends on the forces acting on the object, we must specify precisely what these forces are. In our problem the drag force will at first be assumed to depend on the first power of the velocity and can be written as:
                                                                     F_D= -kv
Where k is a proportionality constant.


1. Draw a detailed motion diagram of the object falling down:

a. Now draw a force diagram for the object falling down. Include vectors for all forces, and write a  statement of Newton's Second. Solve for acceleration.

    a= F / m = D - G = (-kv-mg) / m = - (g + kv / m )


b. Give the condition for the object at terminal velocity (a=0). Using this condition solve for k.

    a= 0, So net force is 0, mg = kv_t , k= mg / v_t


c. Substitute k into your expression for the acceleration from part a.
 
     a = - (g + kv / m )
     k = mg / v_t
     So, a = -g (1+ v / v_t )


2. Open the spreadsheet in the Physics Apps file called air drag. Describe what the spreadsheet is  calculating.

a. What are the assumption? (i.e. initial values? )

     t_o = 0s , g = -9.8m/s^2 , initial velocity = 20m/s , terminal velocity = -40m/s , △t = 0.1s

b. What is v-halfstep?

     V-halfstep is the average velocity in 0.1s.

c. What is the a-halfstep?
 
    A-halfstep is the average acceleration in 0.1s.


3. Using the graph paper provided, draw scales graphs of position vs. time, velocity vs. time and acceleration vs. time for the object no air drag.

position vs. time graph when there is no drag

                                                             p= -4.9t^2 + 20t +1000

velocity vs. time graph when there is no drag

                                                                 v= -9.8t + 20

Make predictions on another sheet of graph paper about how position and velocity graphs would change if you include air drag (D= -kv).

position vs. time graph when drag is equal to -kv

velocity vs. time graph when drag is equal to -kv

Now look at a drag force that is dependent on the square of the velocity. Assuming a drag force,  
F_D= | kv^2 |, find the new formula for the acceleration.

a= F / m = D - G = (kv^2 - mg) / m
mg = k v_t^2 , k = mg / v_t ^2
So, a= [(v^2/ v_t^2) -1] g

position vs. time graph when drag is equal to |kv^2|

velocity vs. time graph when drag is equal to |kv^2|

Conclusion:

     In this lab, we analyzed how changing force affects motion in one dimension. We plotted v-t and p-t graphs in  three cases: with no drag; drag is equal to -kv and drag is equal to | kv^2 |. 

     When there is no drag, the net force on the ball is gravity force, so the acceleration is equal to -9.8m/s^2.  The motion trajectory is parabola. The velocity is changing with constant slope, decreasing to 0 first then increasing in negative position. 

     When drag is equal to -kv, a= -g (1+ v / v_t ). Because the velocity is changing all the time, so the acceleration is also changing. When the ball goes up, velocity is decreasing, so the acceleration will increase in negative position. The maximum height of the ball will be lower than the ball with no drag. Then the velocity will increase in negative position. The velocity will be constant when drag is equal to the gravity force.

     When drag is equal to |kv^2|, according to the v-t graph, we find its acceleration is lower than D= -kv before it reaches the highest point. So, the its maximum height is higher than the ball when D= -kv and D= 0.For balls with  D= -kv and D= |kv^2|, their terminal velocities are the same, and at the highest point their velocities are all equal to 0. According to the v-t graph, when balls are going down the one with D= |kv^2| has a smaller average acceleration. Because v_t^2- v_o^2= 2ax, so the ball with lower acceleration will go deeper. That's why in this case the ball goes deeper than the ball with D= -kv.

      Causes of error:

      We use average velocities in 0.1s to plot our graphs, so the velocities are approximate values, which are not accurate.

         

Lab 7: Centripetal Force

Purpose: To verify Newton's second law of motion for the case of uniform circular motion.

Equipment: Centripetal force apparatus, metric scale, vernier caliper, stop watch, slotted weight set, weight hanger, triple beam balance.

Introduction: Then centripetal force apparatus is designed to rotate a known mass through a circular path of known radius. By timing the motion for a definite number of revolutions and knowing the total distance that the mass has traveled, the velocity can be calculated. Thus the centripetal force, F, necessary to cause the mass to follow its circular path can be determined from Newton's second law.
                                                           F=mv^2/r
Where m is the mass, v is the velocity, and r is the radius of the circular path.
Here we have used the fact that for uniform circular motion, the acceleration, a, is given by:
                                                           a=v^2/r


Procedure:
1. For each trail the position of the horizontal crossarm and the verticle indicator post must be such that the mass hangs freely over the post when the spring is detached. After making this adjustment, connect the spring to the mass and practice aligning the bottom of the hanging mass with the indicator post while rotating the assembly.

2.Measure the time for 50 revolutions of the apparatus. Keep the velocity as constant as possible by keeping the pointer on the bottom of the mass aligned with the indicator post. A while sheet of paper placed as a background behind the apparatus can be helpful in getting the alignment as close as possible. Using the same mass and radius, measure the time for three different trials. Record all data in a neat excel table.

3.Using the average time obtained above, calculate the velocity of the mass. From this calculation the centripetal force exerted on the mass during its motion.

4.Independently determine the centripetal force by attaching a hanging a weight to the mass until it once again is positioned over the indicator post (this time at rest). Since the spring is being stretched by the same amount as when the apparatus was rotating, the force stretching the spring should be the same in each case.

Centripetal apparatus

    Data:
    mass= 475grams
    radius= 16.5cm
                                                                v= 2πr·f                 
                                                                f=50/t
                                                                a=v^2/r
                                                                F=mv^2/r
(r= radius, f= frequency, t= time of 50 revolutions, m= mass, a= acceleration, F=calculated centripetal force)  

The apparatus to find the measured force 

 
 The force diagram for the hanging weight:

Our five trials when m=475grams, r=16.5cm

The average calculated force: 7.36N

The average measured force: 7.066N

Percent difference: 4.44%

5.Add 100g to the mass and repeat steps 2,3,4 above.

    Data:

    mass= 575grams
    radius= 16.5cm

The average calculated force: 7.53N

The average measured force: 7.252N

Percent difference: 3.83%

Conclusion:

      According to our data, we find that when radius is constant, the greater mass, the lower centripetal force. Our average calculated force is really close to measured force, that's means our data is convincing. Because the centripetal force is from the spring. The radius is constant means the length of spring doesn't change, so when the mass changes, the centripetal force is still constant. The formula: F=mv^2/r, F and r are constant, when mass increases, the velocity will decrease.
      In this lab, we verify Newton's second law of motion for the case of uniform circular motion. That the acceleration a of a body is parallel and directly proportional to the net force F acting on the body, is in the direction of the net force, and is inversely proportional to the mass m of the body, F = ma. In this case, the direction of net force points to the center of circle, F= ma, where a=v^2/r. 
      The causes of error:
      (1) When we measure the time for 50 revolutions of apparatus we can't keep the speed exactly constant. So, our frequency is not accurate.
      (2) Our experiment is influenced by air resistance, so our centripetal force actually is not only from the spring.
     

Lab 6: Drag Force on a coffee Filter

Purpose: To study relationship between air drag forces and the velocity of a falling body.


Equipment: Computer with Logger Pro software, lab pro, motion detector, nine coffee filters, meter stick

Introduction: When an object moves through a fluid, such as air, it experiences a drag force that opposes its motion. This force generally increases with velocity of the object. In this lab we are going to investigate the velocity dependence of the drag force. We still start by assuming the drag force,  Fd, has a simple power law dependence on the speed given by Fd=  k |v| ^n, where the power n is to be determined by the experiment.
This lab will investigate drag forces acting on a falling coffee filter. Because of the large surface area and low mass of these filters, they reach terminal speed soon after being released.

Procedure:

NOTE: You will be given a packet of nine nested coffee filters. It is important that the shape of this packet stays the same throughout the experiment so do not take the filters apart or otherwise alter the shape of the packet. Why is it important for the shape to stay the same? Explain and use a diagram.

1. Login to your computer with username and password. Start the Logger Pro software, open the

Mechanics folder and the graphlab file. Don’t forget to label the axes of the graph and create an appropriate title for the graph. Set the data collection rate to 30 Hz.

2. Place the motion detector on the floor facing upward and hold the packet of nine filters at a minimum height of 1.5 m directly above the motion detector. (Be aware other of nearby objects which can cause reflections.) Start the computer collecting data, and then release the packet. What should the position vs time graph look like? Explain.

Verify that the data are consistent. If not, repeat the trial. Examine the graph and using the mouse, select (click and drag) a small range of data points near the end of the motion where the packet moved with constant speed. Exclude any early or late points where the motion is not uniform.

3. Use the curve fitting option from the analysis menu to fit a linear curve (y = mx + b) to the selected data. Record the slope (m) of the curve from this fit. What should this slope represent? Explain.

Repeat this measurement at least four more times, and calculate the average velocity. Record all data in an excel data table.

4. Carefully remove one filter from the packet and repeat the procedure in parts 2 and 3 for the remaining packet of eight filters. Keep removing filters one at a time and repeating the above steps until you finish with a single coffee filter. Print a copy of one of your best x vs t graphs that show the motion and the linear curve fit to the data for everyone in your group (Do not include the data table; graph only please).

5. In Graphical Analysis, create a two column data table with packet weight (number of filters) in one column and average terminal speed (|v|) in the other. Make a plot of packet weight (y-axis) vs. terminal speed not velocity (x-axis). Choose appropriate labels and scales for the axes of your graph. Be sure to remove the “connecting lines” from the plot. Perform a power law fit of the  data and record the power, n, given by the computer. Obtain a printout of your graph for each member of your group. (Check the % error between your experimentally determined n and the theoretical value before you make a printout – you may need to repeat trials if the error is too large.)

6. Since the drag force is equal to the packet weight, we have found the dependence of drag force on speed. Write equation 1 above with the value of n obtained from your experiment. Put a box around this equation. Look in the section on drag forces in your text and write down the equation given there for the drag force on an object moving through a fluid. How does your value of n compare with the value given in the text? What does the other fit parameter represent? Explain.


Our data:

Time vs. Position graph when there is 5 filters

The slope of  Time vs. Position graph represents the terminal velocity (positive direction is upward). So on this graph, the terminal velocity is -1.66m/s. The terminal speed is 1.66m/s.

Time vs. Position graph when there is 8 filters

 On this graph, the terminal velocity is -1.99m/s. The terminal speed is 1.99m/s. 

The average terminal speed of 1-9 trails

We find that when the number of filter decreases, the average terminal speed also decreases. 

                The number of filter vs. average speed graph

We use the power law fit to find the best fitting curve that  Y=2.14X^1.95 (X= terminal speed; Y= number of coffee filters). We find n is 1.95, which is almost 2. That means it matches to the value that given in the text that n=2.

Question:

(1)Why is it important for the shape to stay the same?
     ---Because the drag relates to objects' cross section area. If filter's shape changes, then it's drag will also       change. 

(2)What should the position vs time graph look like?
     ---The curve decreases faster and faster at first. Then, at a moment, the curve turns to be linear, decreases with constant slope.

(3) What should this slope represent (y = mx + b)?
     ---The slope m represents the terminal velocity.

(4)How does your value of n compare with the value given in the text? What does the other fit parameter represent?
    ---We find n is 1.95, which is almost 2. That means it matches to the value that given in the text that n=2. 
         Because Mg= (1/4)Av^2 (A= cross section area). We found that Y=2.14X^1.95 (X= terminal speed; Y= number of coffee filters). So. Y× mg= (1/4)A×(X^2) (m= mass of each coffee filter). Also, Y= [A/ (4gm)]× (X^2). So 2.14 represents A/ (4gm): the cross section area of coffee filter divides four times a coffee filter's gravity force.

Conclusion:

      In this lab, we study the relationship between air drag forces and the velocity of a falling body.  According to our graph and data, we notice that when  an object begins to fall, its speed gradually increases at first. Then, at a moment, the velocity will be constant, and this object will maintain this condition until it toughs down. The reason is that when the speed is increasing, the Drag is also increasing, that Drag=  k |v| ^2. The net force of object is Mg-Drag, so acceleration= (Mg-Drag)/M. The Drag is increasing, the acceleration is decreasing. When Drag is equal to mass of the object, the acceleration turns to be zero. So the object will keep falling will constant velocity.

     The causes of error:
      (1) When filters are falling, their shapes are easy to change. Because drag relates to objects' cross section area, so if filters' shapes change, the drag will also change.
      (2) The filters in the air may be influenced by wind, so they may have extra horizontal velocities. So, the velocity we got may be not correct.  
      (3) We use rounding values to get the "number of filter vs. average speed graph", so it can't be exactly correct.

Lab 5：Working with Spreadsheets

Purpose: To get familiar with electronic spreadsheets by using them in some simple applications.

Equipment: Computer with EXCEL software.

Part 1:

Create a simple spreadsheet that calculates the values of the following function:

f(x)=Asin(Bx+C)

Initially choose value for of A= 5, B= 3 and C= π/3 (1.047).  

Create a column for values of x that run from zero to 10 radians in steps of 0.1 radians. Similarly, create in the next column the corresponding values of f(x) by copying the formula shown above down through the same number of rows (100 in roll).

Our data of two columns

spreadsheet first 20 rows with formulas 

Then copy and paste our data into the graphing program. Put appropriate labels on the horizontal and vertical axes of the graph. Use Curve Fit to find a function that best fit the data.

 The function that best fit the our data

The best fit function: y= 5Sin(3x +1.05)-1.49×10^(-10)

Part 2: 

Repeat the above process for a spreadsheet that calculates the position of a freely falling particle as a function of time. Start off with g= 9.8m/s^2, vo= 50m/s, xo= 1000m and △t= 0.2s.

The formula for free fall: f(t)= ro+vo△t+(1/2)a(△t)^2 

 We assuming the direction of  vo  positive.

(i)When g is positive:

f(t)=1000+ 50t+ (1/2)× 9.8t^2

Our data of two columns when g is positive

Use Curve Fit to find a function that best fit the data:

 The function that best fit the our data when g is positive

The best fit function: f(t)= 4.9t^2+ 50t+ 1000

(ii)When g is positive:

f(t)=1000+ 50t+ (1/2)× (-9.8)t^2

Our data of two columns when g is negative

spreadsheet first 20 rows with formulas when g is negative

Use Curve Fit to find a function that best fit the data:

The function that best fit the our data when g is negative

The best fit function: f(t)= (-4.9)t^2+ 50t+ 1000 

Question: How do data from part 1 and part 2 compare to the values we start with in our spreadsheet?

       In part 1, we compare the data(A=5, B=3, C=1.05) from Curve Fit to the values(A=5, B=3, C=1.047) that we start with in our spreadsheet. In part 2, we do the same thing as initial values are "g= 9.8m/s^2, vo= 50m/s, xo= 1000m and △t= 0.2s". We find that the data from Curve fit are almost the same to the initial values.

Conclusion:
      In this experiment, we try to use electronic spreadsheet to solve problems. We practice the Excel and get familiar to it. In the future, we may use Excel to more and more. So, this experiment is a  good experience to us, that let us know how it works and how it benefits us.
      We find that the data from Curve fit are almost the same to the initial values. This experiment shows that we can utilize Excel and Graphical Fit to collect and summarize the data.
      The causes of error:
       (1) Our data are rounding values so they can't be exactly accurate.
       (2) We collect the data that have the intervals, so our result is an estimate.
     
   

Lab 4: Vector Addition of Forces

Purpose: To study vector addition by:
                          1) Graphical means
                          2) Using components.
              A circular force table is used to check results.

Equipment: Circular force table, masses, massholders, string, protractor, four pulleys.


Part 1:

Our data:

a: 200grams, 0°
b: 100grams, 55°
c: 200grams, 135°

Our scale:  1cm = 20 grams

The magnitudes (length) of vectors:

a: 200grams x (1cm/20grams) =10cm
b: 100grams x (1cm/20grams) = 5cm
c: 200grams x (1cm/20grams) =10cm

Vector diagram:

We find "d" is the resulant force. Then we use ruler and protractor to determine the magnitude (length) and direction (angle) of "d".

The magnitude (length) of "d": 12.5 cm (equal to 250 grams)
The direction (angle) of "d": 62°


Part 2:

Vector components:

a: Ax=10 × cos(0°) =10
    Ay=10 × sin(0°) =0
a= 10i  + 0j

b: Bx=5 × cos(55°) =2.87
    By=5 × sin(55°) =4.1
b= 2.87i + 4.1j

c: Cx=10 × cos(135°) =-7.07
    Cy=10 × sin(135°) =7.07
c= -7.07i + 7.07j

d: Dx= Ax+Bx+Cx =20+5.74-14.14 =5.8
    Dy= Ay+By+Cy =0+4.1+7.07 =11.17
c= 5.8i + 11.17j

The magnitude (length) of "d": √(5.8^2+11.17^2) = 12.59cm (equal to 251.7 grams)
The direction (angle) of "d": arctan(11.17/5.8) ≈62.6°



Part 3:

Mount three pulleys on the edge of force table at the angles. Attach strings to the center ring so that they each run over the pulley and attach to a mass holder. Hang the appropriate masses on each string:


a: 200grams, 0°
b: 100grams, 55°
c: 200grams, 135°

At this moment, the ring is not equilibrium.

Set up a fourth pulley and mass holder at 180 degrees opposite from the angle you calculated for the resultant of the first three vectors.

d: 251.7 grams; 62.6°+180° =242.6°

When we place a mass on fourth holder equal to the magnitude of the resultant, the ring turns to equilibrium.

The ring is in equilibrium after we the fourth mass

Part 4:

We confirmed our result via simulation:

The simulation of our data

We found the simulation data of the resultant are really close to our resultant data (62.6°), that means our data are convincing.

Conclusion:

     When we place a mass on fourth holder equal to the magnitude of the resultant, the ring turns to equilibrium. That means the force of the fourth mass is equal to the resultant  force of the first three masses.
     This experiment proved that force has direction, and a resultant force consists several vector forces. A vector is a quantity having a magnitude and a direction, and two vectors of the same type can be added.
      The sources of error:
            1. Some magnitude of vectors are decimals, but we only have the masses with whole numbers.
            2. Some masses are rusted, so their mass may be higher or lower than the standard.
            3. Our table is not horizontal, so our directions of vectors are little biased.

Lab 3: Acceleration of Gravity on an Inclined Plane

Purpose: 1. To find the acceleration of gravity by studying the motion of a cart on an incline.
                2. To gain further experience using the computer for data collection and analysis.

Equipment Needed: Windows based computer with Logger Pro software, motion detector, ballistic cart,  aluminum track, wood blocks, meterstick, small carpenter level.

Introduction: 
      In this laboratory you will use the computer to collect position (x) vs time (t) data for a cart accelerating on an inclined track . By comparing the acceleration of the cart when moving up and down the track, the effect of friction can be eliminated and the acceleration due to the effect of gravity alone can be found.
      Since the force of friction acts with the force of gravity when the cart is going up the track and against the force of gravity when the cart is going down the track, we can average the slightly increased acceleration (when going up) with the slightly decreased acceleration (down) to obtain an acceleration that depends only on the force of gravity. If we call g the acceleration due to gravity when an object is in free fall, then the component of this acceleration along the track is gsinθ where θ is the angle of inline for the track. Thus:

                                                               gsinθ=(a1 + a2) / 2

Where a1 and a2 are the acceleration of the cart up and down the inline. In this lab we will measure acceleration by looking at the slope of the v vs t curve for the cart.


Analysis:

When cart goes up, the force of friction is negative (we assume that the direction of velocity is positive). If the acceleration of friction is f, then:
                                                                 a1 =  | gsinθ | + | f |

When cart goes down, the force of friction is positive (we assume that the direction of velocity is negative). If
the acceleration of friction is f, then:
                                                                  a2  =  | gsinθ | - | f |

So: (a1 + a2) / 2 =（ | gsinθ | + | f |  + | gsinθ | - | f | ）/ 2 = (2 | gsinθ |) / 2 = | gsinθ |



Determine the θ:

a & b: The distance between aluminum track's endpoints to the desktop.
c: The length of the aluminum track.

In order to get the inclination angle θ, we measured the length of a,b,c:
a=6.3cm
b=13.2cm
c=227cm

so:
d=a-b=13.2cm-6.3cm=6.9cm
sinθ=d/c=6.9cm/227cm≈0.03
θ=arcsin(0.03)≈1.74°


Part 1:

R-T and V-T graphs 

According to our position vs. time graph (position is the distance from detector to cart) :
Position is decreasing between 0.6-3.2s, so the cart is going up;
Position is increasing between 3.2-5.8s, so the cart is going up.

On velocity vs. time graph:
 Velocity is negative between 0.8-3.2s, the cart is going up:
 Velocity is positive between 3.2-5.8s, the cart is going down:

Because the acceleration is the slope of velocity vs. time graph, we use the linear function to fit the curve and find the slopes when cart goes up and cart goes down.

When cart goes up:
a1 = 0.3 m/s^2

When cart goes down:
a2 = 0.25m/s^2

So:
                                                          gsinθ = (a1 + a2) / 2
                                                     g × 0.03 = (0.3m/s^2+0.25m/s^2)/2
                                                     g × 0.03 = 0.275m/s^2
                                                                g = 9.2m/s^2


Then we repeat our experiment two more times to get the average data:

Our gravity data when  sinθ=0.03

Part 2:

Then we use a lager block of wood to increase the angle of inclination of the track:

a=5.1cm
b=19.5cm
c=227cm

so:
d=a-b=19.5cm-5.1cm=14.4cm
sinθ=d/c=14.4cm/227cm≈0.06
θ=arcsin(0.06)≈3.64°

Our gravity data when  sinθ=0.06

Conclusion:
  According to two experiments, we found when θ is larger, our experimental data are closer to actual data(0.5% diff compare to 8.2% diff). The reason is that when θ is larger, the motion of cart is closer to free fall, which is influenced less by the disturbance. 
The causes of error: 
(1) Air resistance also against the motion. 
(2) Our table is not horizontal, so our θ is not precise enough. 
(3) The error of the equipment and the error when we read the data.
In this lab, we learned acceleration along the track is gsinθ where θ is the angle of inline for the track. we can use this property to estimate the gravity. We also learned how to control the variable to get another group of data, then try to think abut what cause the difference.

Lab 2：Acceleration of Gravity

Purpose: 1. To determine the accelaration.
                2. To gain experience using the computer as a data collector.

Equipment: Windows based computer, Lab Pro interface, Logger Pro Software, motion detector, rubber         ball, wire basket.

Introduction: In this laboratory you will use the computer to collect some position (x) vs time (t) data for a rubber ball tossed into air. Since the velocity of an object is equal to the slope of the x vs t curve, the computer can also construct the graph of v vs t by calculating the slope of x vs tat each point in time. We will  use both the x vs t graph and the v vs t graph to find the free fall acceleration of the ball.

Part 1:

  (1) Position vs. time graph:

       position (x) [m] vs. time (t) [s] graph

  (2) Function:

      We selected an appropriate data range and tried to find a function that fit the curve. 

      We chose the Quadratic (At^2+Bt+C), our equation is "x = -4.858t^2 + 8.535t - 2.163".

      (A = -4.858,  B = 8.535,  C = -2.163)

  (3) Slope:

      Because "velocity = (△x) / (△t)" , which is the derivative of the position (x) vs. time (t) curve.
      So the slope of the  position vs. time curve is the velocity.

  (4) Velocity:

      According to the graph, we found that between 0.5s and 0.85s the slope is positive, if we assume the upward direction is positive direction, velocity between 0.5s and 0.85s is positive; the slope between 0.85s and 1.33s is negative, so the velocity is also negative.

  (5) Acceleration:

     For a linear motion with constant acceleration: during the time interval "△t = tf - ti " ,

     sf = si+vis△t+(1/2)as(△t)^2 ——  note that "sf = final position", "si = initial posotion", 

                                                                      "vis=initial velocity", "as = accelaration".

   
     So "A" in our quadratic equation is equal to (1/2) as .      

     So  acceleration=2A=2 x (-4.858)= -9.716m/s^2

  

  (6) Gravity:   

     We assume that gravity is equal to acceleration.
     So our "gexp =2A=2 x (-4.858) =-9.716m/s^2 ".

     (Accepted value of gravity: gacc= -9.8m/s^2)

  (7) Error: 

     Percent error = [(measured- actual) / (actual)] x 100%  

                          =  [(9.8m/^2-9.716m/^2) / (9.8m/s^2)] x 100% = 0.92% .

     That means our data is really close to the accepted value.

Part 2:

     

 (1) Velocity vs. time graph:

    velocity (v) [m/s] vs. time (t) [s] graph

  
  (2) Function: 

          We selected an appropriate data range and tried to find a function that fit the curve. 

          We chose the linear function (v=mt+b), our equation is "v = -9.799t + 8.613".

          (m = -9.799, b = 8.613)

   
  (3) Slope:

         Because "acceleration = (△v) / (△t)" , which is the derivative of the velocity (v) vs. time (t) curve.
         So the slope of the velocity vs. time curve is the acceleration.

   
  (4) Acceleration:

         Because  the slope (m) of the velocity vs. time curve is the acceleration. 

         So the "acceleration = m = -9.799 m/s^2".

  (5) Gravity:

        We assume that gravity is equal to acceleration, so our "gexp = m = -9.799 m/s^2". 

        (Accepted value of gravity: gacc= -9.8m/s^2) 

 
  (6) Error: 

       Percent error = [(measured- actual) / (actual)] x 100%  

                          =  [(9.8m/^2-9.799m/^2) / (9.8m/s^2)] x 100% = 0.01% .

       That means our data is really close to the accepted value.

Then we repeated our experiment for several times to get the average data:

Conclusion:

       In this lab, we determine the gravity is close to 9.8m/s^2.
       In order to get the most precise data, we did this experiment for many times and got the average data, which decrease the error accidental error. Our data is 3.18% varying from the actual, that mean this experiment can prove that the gravity for a freely falling object is close to 9.8m/s^2. Our errors are because of :
      1. Air resistance
      2. The inevitable experimental error, because the equipment can't be exactly precise.
      3. The curve fit is an estimate, so our gravity is also an estimate value.

      We also learned how to use Lab Pro interface, Logger Pro Software, motion detector and gained the experience using the computer as a data collector. We also worked as a team to gain and analysis the data. This experience may benefit us in the future.