Physics Lab 1 Position and Velocity PDF

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Physics 110 Lab 1 online...

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Introductory Physics Hunter College Position and Velocity Abstract: Th ep o s i t i o no fa no b j e c ta t ap a r t i c u l a rmo me n ti nt i mei sf u n d a me nt a l f ord e t e r mi n i n g v e l oc i t yi n1 d i me n s i on a lmo t i o nwi t h o uta c c e l e r a t i o n . Asa no b j e c tmo v e sa l on gap a t h ,i t a c c u mu l a t e sama g n i t u d eofd i s t a n c e . Di s p l a c e me n t ,h o we v e r ,i st h ed i ffe r e n c eb e t we e nt h e s t a r t i n gp o i n ta n de n dp o i n twi t ht h ei n c l u s i ono fd i r e c t i o n . Sp e e di same a s ur e me n to fd i s t a n c e o v e rac e r t a i na mou n tt i mea n dv e l o c i t yi same a s ur e me nt o fd i s p l a c e me n to v e rac e r t a i na mo u n t o ft i me .Sp e e da n dd i s t a n c ea r es c a l a rq ua n t i t i e st h a to n l yi n c l u dema g n i t u d e s , b u tv e l o c i t ya n d d i s p l a c e me nta r ev e c t o rqu a n t i t i e sa n di n v o l v ed i r e c t i o na l o n gwi t hma g n i t u d e . Th r ou g hg r a p h i c ma n i p u l a t i o no fp o s i t i onwi t ht i me , a v e r a g ev e l o c i t i e sa n da v e r a g es p e e d swe r ed e t e r mi n e df r o m d i s p l a c e me nt sa n dd i s t a n c et r a v e l l e d .Ve l o c i t ywa sd e t e r mi ne dgr a p h i c a l l yf r o mt h es l o p ea n d t )=vt+x0 wa sc on fir me da l g e b r a i c a l l yt h r o ug ht h ee q u a t i on x 2 c a l c( Objectives: Students will learn how position relates to constant velocity by predicting positions and velocities of given position vs. time and velocity vs. time graphs. Students will also calculate the slope of position vs. time graph to calculate the velocity. Finally, students will determine position from the equation x(t) = v 0 t + x0 . Background Any measurement of position, distance, or speed must be made with respect to a reference frame. For example, if you are sitting on a train and someone walks down the aisle, their speed with respect to the train is a few miles per hour, at most. Their speed with respect to the ground is much higher. For today’s simulation we will use the cartesian coordinate plane, xy-axis, to orient our frame of reference. Distance vs. Displacement Displacement (blue line) is how far the object is from its starting point, regardless of how it got there. Displacement depends on direction and thus, is a vector quantity; it can be a negative, or positive value that indicates its direction and magnitude. The displacement is written: ∆x = x2 − x1 Distance traveled (dashed line) is measured along the actual path. Distance is a length, and is thus a scalar quantity. Distance can be any measure x ≥ 0 units.

Average Speed vs. Velocity Page 1 of 9

Speed is how far an object travels in a given time interval

Velocity is speed and includes directional information.

For constant velocity, non-accelerated motion, we can use the equation v = ∆x/∆t to solve for either position, or velocity, or time, if we know the other quantities .

Pre-Lab Questions

1. Consider a deer that runs from point A to point B. The distance the deer runs can be greater than the magnitude of its displacement, but the magnitude of the displacement can never be greater than the distance it runs. a) True b) False 2. Consider a car that travels between points A and B. The car's average speed can be greater than the magnitude of its average velocity, but the magnitude of its average velocity can never be greater than its average speed. a) True b) False 3. Complete the partially filled in table below Physical Quantity Scalar

Vector

Position

Distance

displacement

Motion

Speed

Velocity

Time

Seconds/hours/minutes

NA

Procedures Getting Started 1. Go to http://physics.bu.edu/~duffy/HTML5/1Dmotion_graph_matching.html We will use this Physics Simulation to practice with non-accelerated motion in 1D. Your screen should look like the image below:

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2. Please look at the position vs time, velocity vs time graphs, and the motion diagram, which shows the displacement of motion of the red dot. 3. Notice there are Play, Pause, >, and Reset tabs you will use throughout this simulation to modify your simulation as it plays. Click Pause, > , Reset to change the variables, as necessary. 4. There are 3 sliders that control Initial position (from -50m to +50m), Velocity (from -10m/s to +10m/s), and Acceleration (from -2.0 m/s/s to 2.0 m/s/s). We will keep the acceleration at zero for all of today’s simulations. 5. Under the 3 sliders there are Match a position graph and Match a velocity graph . Today we will use Match a position graph. Graph Matching 6. Review the 5 different Position vs. time and velocity vs. time graphs below.

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7. Predict the velocity that will determine the position for each of the 5 Position vs. time graphs using the controls on the simulation, explained in the Getting Started steps. 8. Go to the simulation Matching Graphs. 9. Click Graph 1, at the bottom of the screen, . Predict how to match this graph by predicting how you should choose the initial position and the velocity values with the sliders at time t = 0 s, 10 s, & 20 s. Note: Each of these graphs may have one to three different values for position and velocity. 10.Press the Play tab to run the simulation, with your predictions. 11.Did you match the graph perfectly? If not, change the Initial position and velocity controls until you have matched the graph. Then record the data from the graphs into the Matching Graphs table below. You may need to pause the simulation to change the velocity at certain times. Use > tabs to slowly increment the motion in time steps of 0.05s. 12.Repeat steps 8 – 10, for Graphs 2 – 5. 13.Calculate distance, displacement, average speed, and average velocity for each of the 5 graphs you matched; enter your results in the table Matching Graphs. Table: Matching Graphs Graph One Time t (s)

Position x(t) (m)

Velocity (m/s)

t1=0

x1= 50

v1= 0

t2=20

x2=50

v2=0

t3=

x3=

v3=

Total Distance (m)

Total Displaceme nt (m)

Average Speed (m/s)

Average Velocity (m/s)

0s

0s

0 m/s

0 m/s

Average Speed

Average Velocity

Graph Two Time t (s)

Position x(t) (m)

Velocity (m/s)

Total Distance

Total Displaceme

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t1=0

x1=0

v1=5

(m)

nt (m)

(m/s)

(m/s)

t2=20

x2=100

v2=5

100m

100m

5 m/s

5 m/s

t3=

x3=

v3= Graph Three

Time t (s)

Position x(t) (m)

Velocity (m/s)

t1=0

x1=-50

v1=0

t2=10

x2=-50

v2=+10

t3=20

x3= +50

v3= +10

Total Distance (m)

Total Displaceme nt (m)

Average Speed (m/s)

Average Velocity (m/s)

100m

100m

5 m/s

5 m/s

Graph Four Time t (s)

Position x(t) (m)

Velocity (m/s)

t1=0

x1=0

v1=5

t2=10

x2=50

v2=5

t3=20

x3=-50

v3=-10

Total Distance (m)

Total Displaceme nt (m)

Average Speed (m/s)

Average Velocity (m/s)

150m

-50m

7.5 m/s

-2.5

Graph Five Time t (s)

Position x(t) (m)

Velocity (m/s)

t1=0

x1=50

v1=-3

t2=10

x2=20

v2=-3

t3=20

x3=20

v3=0

Total Distance (m)

Total Displaceme nt (m)

Average Speed (m/s)

Average Velocity (m/s)

30m

-30m

1.5 m/s

-1.5 m/s

Questions 14.Is there any instance where the total displacement is greater than total distance? Explain. There should not be any instance where total displacement is greater than total distance. Distance is a scaler quantity that must contain the sum of all magnitudes without taking account direction. Displacement is a vector quantity and takes direction into account, therefore making it either the same as distance or less than total distance as seen in the data above. If a position value is negative, that means the object is moving backwards and that will affect overall displacement of the object.

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15.Is there any instance where the average velocity is greater than the average speed? Explain. There is also no instance where average velocity is greater than speed. Since velocity is dependent on displacement and displacement can never be greater than the total distance, then the average velocity has to be either the same or less than the average speed. Also, speed is dependent on distance and since distance can be equal or greater than displacement, speed will be equal or greater than velocity.

Motion with Constant Velocity: 16.Now play the simulation, without matching, by selecting the tab at the bottom of the screen, just below the match graph options. 17.Move the Initial position slider to 0m. 18.Move the Velocity slider, to choose any velocity between 1 and 10 m/s. 19.Set the acceleration slider to 0m/s2. 20.Now run the simulation to see how long it will take. 21.Divide the total time of your simulation by 4. 22.Play your simulation again, so that you may now record the position and time for the 23.first quarter of the simulation in the Constant Velocity table below. 24.Continue the simulation until you get to the 2nd quarter, 3rd quarter, and the end of your simulation and record those positions and times in the Constant Velocity table. 25.Take a screenshot of your position graph, after simulation runs completely, and enter it here, like this one:

26.Calculate 2 slopes. The first between 0 and the first quarter of your simulation and the second from third quarter to the end of your simulation. Page 6 of 9

27.Enter your slope calculations in the Constant Velocity table below:

Table: Constant Velocity Velocity [m/s] = 5m/s Data Collected

Calculate

Time [s]

Position [m]

0

t0 = 0

x0 = 0

1

t1 = 5

x1 =25

2

t2 = 10

x2 =50

3

t3 = 15

x3 =75

4

t4 =20

x4 =100

Slope= Δx/Δt [m/s]

(Δx/Δt)1 = 5m/s

(Δx/Δt)2 = 5m/s

(Δx/Δt)avg = 5+5/2 = 5m/s

28.Compare the average slope with the velocity, i.e. calculate the percent error between the velocity and your calculated average slope. % error = 0% ((5m/s-5m/s)/5m/s)x100 = 0% 29.Now calculate the position, x2calc, value using your recorded data from the Constant Velocity table and the equation x2calc(t) = vt + x0. Enter your results in the Position Data table below: Table: Position Data Graph Position Xgraph

Calculated Position Xcalculated

Percent Error Xcalculated vs. Xgraph

X2graph 20m

x2calc (5m)(4s)+0m = 20m

0%

30m

(5m)(6s)+0m = 30m

0%

60m

(5m)(12s)+0m = 60m

0%

30.Compare x2calc with the value x2 from your data, i.e. calculate the percent error, for your calculated position vs. your measured position. 31.Repeat steps 21 for 2 more positions; choose any 2 positions you like. Post-Lab Questions

1. The figure shows a graph of the position of a moving object as a function of time. What is the velocity of the object at each of the following times? a) At t = 1.0 s = 10m/s 20m-0m/2s-0s = 10 m/s b) At t = 2.5 s = 20m/s 40m-20m/3s-2s = 20m/s c) At t = 4.0 s= 0 m/s slope=0 d) At t = 5.5 s= -40 m/s -40m-0m/6s-5s = -40 e) What is the average velocity of the object from t = 0 s to t = 4.0 s? Page 7 of 9

(40m-0m)/ (4s-0s)= 10m/s f) What is the average velocity of the object from t = 0 s to t = 6.0 s? (0m-0m)/ (6s-0s)= 0m/s

2. The graph in the figure shows the position of an object as a function of time. The letters H-L represent particular moments of time.

a) At which moment in time is the speed of the object the greatest? J (steepest slope) b) At which moment in time is the speed of the object equal to zero? I slope=0 3. If you run a complete loop around an outdoor track of length 400 m in 100 s, find your a) average velocity Displacement = 0m xfinal=xinitial. 0m/100s = 0m/s b) average speed. Speed= 400m/100s= 4m/s 4. A polar bear starts at the North Pole. It travels 1.0 km south, then 1.0 km east, and then returns to its starting point. This trip takes 0.75 hr. a) What was the bear's average speed? 1km+1km+ total 3.41km/0.75hr= 4.55 km/hr

√ 2 = 3.41 km

b) What was the bear's average velocity? Total displacement = km, √ 2 km/ .75hr= 1.89m/s 5. Two locomotives 70 kilometers apart are travelling on the same track towards each other , Engine A moves at 22 kilometers per hour east and engine B moves at 13 kilometers per hour west. At the instant both trains begin moving, an annoying mutant fly begins flying from engine A towards engine B at 33 kilometers per hour . The instant it touches Page 8 of 9

B, it turns around and flies back. It goes on this way until the two locomotives collide and the mutant fly is finally squashed.

So, before its untimely demise, determine the following: a. the total distance the fly flew t= 70 km / (22km/hr+13km/hr) = 2 hours 33km/hr x 2 hr = 66km b. the time it took till it was eliminated 2 hours c. The average velocity of the fly

Distance 1 = (70km/ (13 km/hr +33 km/hr)) x 33 km/hr = 50.22 km. (70km/ (13 km/hr +33 km/hr)) x 22 km/hr = 33.5 km Distance 2 = (50.22km – 33.5 km) / (33 km/hr + 22 km/hr) =7km Distance 3: (7.0 km - 2.8km) / (13 km/hr +33 km/hr) x33 km/hr = 3km

(50.22 km - 7km + 3km) / 2hr = 23.1 km/hr

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