# The Moving Man

```Kinematics
By
Elenis Vergara Martinez
Abstract
In this lab, I set out to interpret, predict, and draw charts of position, velocity and
acceleration for common situations by using the phet lab platform on Kinematics
site.) By providing different scenarios of motion I was able to become more familiar
with the concepts of displacement, velocity and acceleration. Also, the creation of
graphs, movement diagrams and realization of kinematics problems clarified my
knowledge regarding this topic
Introduction
The objective of this laboratory is to explore velocity, and
acceleration graphs, as demonstrated by a moving man. Will
be used an online simulation (screen shown in Figure 1) from
the University of Colorado Boulder called, PHET Simulation –
Moving Man.
This simulation allows you to explore topics related to the
velocity, position, and acceleration of an object. In order to be
able to relate-predict motion graphs for common random situations.
Figure 1
In this laboratory, several scenarios of constant speed and accelerated movement
will be analyzed and then the results obtained will be valued through the use of
kinematic equations. Consequently, different predictions of graphs, questions and
problems will be proposed in order to delve deeper into the main topic.
On the straight line we place an origin O, where an observer will be who will measure
the position of the mobile x at instant t. The positions will be positive if the mobile is
to the right of the origin and negative if it is to the left of the origin. The position x of
the mobile is related to time t by means of a function x=f(t). Suppose now that at time
t, the mobile is in position x, later, at time t’ the mobile will be in position x’. We say
that the mobile has moved Δx=x’-x in the time interval Δt=t’-t, measured from instant
t to instant t’. To determine the velocity at time t, we must make the time interval Δt
as small as possible, in the limit as Δt approaches zero. In general, the speed of a
body is a function of time. Suppose that at an instant t the speed of the mobile is v,
and at instant t’ the speed of the mobile is v’. Average acceleration between instants
t and t' is the ’uotient between the change in’speed Δv=v’-v and the time interval in
which it has taken to make said change, Δt=t’-t.
The kinematic formulas are a set of formulas that relate the five kinematic variables
listed below:
Δx displacement
T time interval
V0 initial velocity
V final velocity
a accelertion
Data Table 1 – Constant velocity
This first table shows the results of the final position and time of the moving man
simulation in different cases.
a. The moving man walking away from the origin to the house slowly and steadily.
(Steadily means with a constant speed.) To the house
b. The moving man walking away from the origin medium fast and steadily to the
tree.
c. The moving man walking toward the house slowly and steadily from the tree.
d. The moving man walking toward the tree medium fast and steadily from the house.
Velocity (m/s)
Time (s)
Initial Position
Final Position
(m)
(m)
Trial A
1.4
5.8
0.0
8.1
Trial B
-2.5
3.3
0.0
-8.1
Trial C
1.4
11.5
-8.0
8.1
Trial D
-2.4
6.7
8.0
-8.1
* Humans tend to walk at about 1.4 meters per second or 5 km per hour. Although
humans are capable of walking at speeds up to 2.5 m per second or 9 km per hour at
level, humans typically choose to use only a small range within these speeds. (Zyl)
Data Table 2 – Accelerated Motion
This first table shows the results of the experimental displacement, experimental
velocity and time of the moving man simulation for each of the given situations.
a. The moving man starts from the house and moves towards the left with a constant
acceleration at a slow speed.
b. The moving man starts near the right wall (10 m) with a velocity of 0 m/s and
acceleration of – 1.5 m/s2 to the tree.
c. The moving man starts near the left wall (-10 m) with a velocity of 4.5 m/s and an
acceleration of -0.45 m/s2 to the house.
d. The moving man walking away from the origin with a velocity of 2.5 m/s and an
acceleration of -0.35 m/s2 to the house during 12.6 s.
Acceleration
(m/s2)
Time (s)
Initial
Position (m)
Trial A
Initial
Velocity
(m/s)
-1.4
8.0
Experimental
Displacement
(m)
0.125
Experimental
Final Velocity
(m/s)
-1.4
0.0
5.6
Trial B
0.0
-1.5
5.0
10.0
-8.4
-7.4
Trial C
4.5
-0.45
5.7
-10.0
-8.3
1.95
Trial D
2.5
-0.35
12.6
0.0
3.7
-1.9
Data analysis
Constant velocity
Calculations of each table result.
a. The moving man walking away from the origin to the house slowly and
Trial A
Velocity (m/s)
Time (s)
1.4
5.8
Initial Position
(m)
0.0
Final Position
(m)
8.1
Figure 1. Description of the motion of the moving man for the scenario A from the
starting point
Figure 2. Final position of the moving man for the scenario A.
Calculated displacement
𝑋 − 𝑋0
𝑉=
→ 𝑋 = 𝑋0 + 𝑣𝑡 = ( 0.0) + (1.4)(5.8) = 8.1𝑚
𝑡
b. The moving man walking away from the origin medium fast and steadily to
the tree.
Trial B
Velocity (m/s)
Time (s)
-2.5
3.3
Initial Position
(m)
0.0
Final Position
(m)
-8.1
Figure 3. Description of the motion of the moving man for the scenario B from the
starting point.
Figure 4. Final position of the moving man for the scenario A.
Calculated displacement
𝑋−𝑋
𝑉 = 𝑡 0 → 𝑋 = 𝑋0 + 𝑣𝑡 = ( 0.0 ) + (−2.5)(3.3) = − 8.2m
c. The moving man walking toward the house slowly and steadily from the tree.
Trial C
Velocity (m/s)
Time (s)
1.4
11.5
Initial Position
(m)
-8.0
Final Position
(m)
8.1
Figure 5. Description of the motion of the moving man for the scenario C from the
starting point
Figure 6. Final position of the moving man for the scenario C.
Calculated displacement
𝑋−𝑋
𝑉 = 𝑡 0 → 𝑋 = 𝑋0 + 𝑣𝑡 = (−8.0 ) + (1.4)(11.5) =8.1m
d. The moving man walking toward the tree medium fast and steadily from the
house.
Trial D
Velocity (m/s)
Time (s)
-2.4
6.7
Initial Position
(m)
8.0
Final Position
(m)
-8.1
Figure 7. Description of the motion of the moving man for the scenario D from the
starting point.
Figure 8. Final position of the moving man for the scenario D.
Calculated displacement
𝑋 − 𝑋0
𝑉=
→ 𝑋 = 𝑋0 + 𝑣𝑡 = (8.0 ) + (−2.4)(6.7) = −8.08 ≈ −8.1𝑚
𝑡
Accelerated motion
Calculations of each table result
a. The moving man starts from the house and moves towards the left with a
constant acceleration at a slow speed.
Trial A
Initial
Velocity
(m/s)
-1.4
Acceleration
(m/s2)
Time (s)
0.0
5.6
Initial
Position
(m)
8.0
Experimental
Displacement
(m)
0.125
Experimental
Final Velocity
(m/s)
-1.4
Figure 9. Description of the motion of the moving man for the scenario A from the
starting point.
Figure 10. Displacement of the moving man for the scenario A.
Calculated displacement
1
𝑋 = 𝑋0 + 𝑉0 𝑡 + 𝑎𝑡 2
2
𝑋 = (8.0) + (−1.4)(5.6) + 1/2(0.0)(4.9)2
X= 0.16
Calculated Velocity
𝑉 = 𝑉0 + 𝑎𝑡
𝑉 = (-1.4) + (0.0) (5.6) = -1.4 m/s
b. The moving man starts near the right wall (10 m) with a velocity of 0 m/s and
acceleration of – 1.5 m/s2 to the tree.
Trial B
Initial
Velocity
(m/s)
0.0
Acceleration
(m/s2)
Time (s)
Initial
Position (m)
-1.5
5.0
10.0
Experimental
Displacement
(m)
-8.75
Experimental
Final Velocity
(m/s)
-7.5
Figure 11. Description of the motion of the moving man for the scenario B from the
starting point
Figure 12. Displacement of the moving man for the scenario B.
Calculated displacement
1
𝑋 = 𝑋0 + 𝑉0 𝑡 + 𝑎𝑡 2
2
𝑋 = (10.0) + (0.0)(5.0) + 1/2(−1.5)(5.0)2
X= -8.75 m
Calculated Velocity
𝑉 = 𝑉0 + 𝑎𝑡
𝑉 = (0.0) + (-1.5) (5.0) = -7.5 m/s
c. The moving man starts near the left wall (-10 m) with a velocity of 4.5 m/s and
an acceleration of -0.45 m/s2 to the house.
Trial C
Initial
Velocity
(m/s)
4.5
Acceleration
(m/s2)
Time (s)
Initial
Position (m)
-0.45
5.5
-10.0
Experimental
Displacement
(m)
-8.02
Experimental
Final Velocity
(m/s)
2.006
Figure 13. Description of the motion of the moving man for the scenario C from the
starting point.
Figure 14. Displacement of the moving man for the scenario C.
Calculated displacement
1
𝑋 = 𝑋0 + 𝑉0 𝑡 + 𝑎𝑡 2
2
𝑋 = (−10.0) + (4.5)(5.5) + 1/2(−0.45)(5.5)2
X= 7.94 m
Calculated Velocity
𝑉 = 𝑉0 + 𝑎𝑡
𝑉 = (4.5) + (-0.45) (5.5) = 2.02 m/s
d. The moving man walking away from the origin with a velocity of 2.5 m/s and an
acceleration of -0.35 m/s2 to the right during 14.2s. Watch his motion.
Trial D
Initial
Velocity
(m/s)
2.5
Acceleration
(m/s2)
Time (s)
Initial
Position (m)
-0.35
14.2
0.0
Experimental
Displacement
(m)
0.089
Experimental
Final Velocity
(m/s)
-2.48
Figure 15. Description of the motion of the moving man for the scenario D from the
starting point.
Figure 16. The moving man arrives to the house and stopped about 2 seconds.
Figure 17. The moving man moves back until he arrives to the origin.
Calculated displacement
1
𝑋 = 𝑋0 + 𝑉0 𝑡 + 𝑎𝑡 2
2
𝑋 = (0.0) + (2.5)(14.2) + 1/2(−0.35)(14.2)2
X= 0.21m
Calculated Velocity
𝑉 = 𝑉0 + 𝑎𝑡
𝑉 = (2.5) + (-0.35) (14.2) = -2.47 m/s
Discussion and Conclusion
During the laboratory the relationship between the speed, acceleration, time and
position of an object was studied with the help of the simulation of the moving man.
The laboratory was carried out under different cases related to speed, time,
acceleration in the individual. Therefore, the use of certain kinematic equations was
required for the valorization of the experimental data.
The simulation was carried out under the close approximation of the ideal frictionless
motion. The initial velocities, position and acceleration of the man in certain time
intervals are measured for different situations detailed in the report. And from the
data obtained, the final displacement and the final speed of the object are calculated
using the equations mentioned in the introduction. Certain results ended up with a
tiny little decimal difference. However, they were different due to the number of
decimal places obtained in the simulation.
Finally, certain graphs are presented representing each of the results obtained from
the different scenarios. Thus, illustrating the different relationships between position,
velocity, time and acceleration. Furthermore, this gives the possibility of answering
questions related to the subject under investigation. Providing examples,
calculations and illustrations. The results obtained in the experiment are duly
reasonable, but there may be errors due to misreading the data or due to imprecision
when performing the calculations.
1. Predict how the position and velocity graphs would look for the scenario A of the
moving man having constant speed. Give a brief explanation.
A. The moving man walking away from the origin to the house slowly and
As we see, man moves away from the origin. This is indicated by the blue increasing
straight line in the positive quadrant of the position versus time graph. Also, his
movement is constant, so the speed didn’t change at all. Therefore, the slope of the
graph of velocity vs. time is zero, as is the acceleration of it.
2. Present a basic motion diagram for the scenario C od the moving man having
constant speed. Give a brief explanation.
d. The moving man walking toward the house slowly and steadily from the
tree.
→
V= 1.4
The velocity vector is pointing to the right because the object is moving in the positive
direction.
3. What characterizes constant-velocity motion on a position-time graph?
The principle is that the slope of the position-time graph gives a lot of useful
information. If the objects move at a constant speed, then the slope is constant (a
straight line). If the velocity changes during the motion, then the slope changes (a
curved line)
For instance, in scenario B of constant velocity, the position-time graph should look
like this.
b. The moving man walking away from the origin medium fast and steadily
to the tree.
As the object is moving at a constant speed, the
slope of the graph is zero.
4. What is the area under a speed-time graph equal to? Represent your answer
with an example.
The area under a speed-time graph is the distance travelled in a certain amount of
time.
The area of the rectangle is found by:
A= b x h = (3)(6) = 18
The area of the triangle is found by:
A= (b x h)/2 = (4)(6)/2 =12 m
Total area =Total displacement
= 18 + 12 = 30m
5. What does the slope represent in the velocity-time graph?
The slope in the velocity-time graph represents the acceleration of the object.
The formula of the slope of a velocity graph is:
Slope=Δt/Δv
and
Let&acute;s look an example of a velocity graph of the
moving man simulation.
The slope of the curve is positive between t=0s and
t=2s since the slope is directed upward. This means
the acceleration is positive.
The slope of the curve is negative between t=2s and t=4.2s since the slope is
directed downward. This means the acceleration is negative.
6. Predict how the position, velocity, and acceleration graphs would look for the
scenario D of the moving man having accelerated motion. Describe the motion.
D. The moving man walking away from the origin with a velocity of 2.5 m/s and
an acceleration of -0.35 m/s2 to the house during 12.6 s.
The moving man moves away from the origin. This is indicated by the blue increasing
curved line in the positive quadrant of the position versus time graph. Note that the
motion is moving like a curve. Therefore, there is a change in velocity (acceleration).
The object has a positive velocity with a negative acceleration, so the object is
slowing down.
At t= 7s the man isn’t moving and then he goes back to the origin speeding up in
negative direction.
7. What does the area represent on a acceleration vs. time graph?
The area under the acceleration vs. time graph represents the change in velocity
area=Δv
For instance,
The area of the triangle is found by:
A= (b x h)/2 = (8)(6)/ 2 = 24 m/s
This is in velocity during the given time interval. For us to get
the final velocity, we need to use the definition of change in
velocity.
Δv=vf−viΔv=vf−vi
Vf= 24 + 20 = 44m/s
8. Present a basic motion diagram for the scenario C od the moving man having an
accelerated speed. Give a brief explanation.
c. The moving man starts near the left wall (-10 m) with a velocity of 4.5
→
V= 4.5
m/s and an acceleration of -0.45 m/s2 to the house.
0m
The velocity vector is pointing to the right because the object is moving in the positive
direction while slowing down due to its negative acceleration.
→
A= -0,45
9. The man accelerates uniformly from rest to a speed of 3.2 m/s over a distance
of 8.6 m. Determine the acceleration of the moving man. Present the graphics.
vf2 = vi2 + 2a∆𝑥
(3.2 m/s)2 = (0 m/s)2 + 2(a)(8.6 m)
10.2m2/s2 = (0 m/s)2 + (17.2 m)a
(9.61m2/s2 )/(17m) = a
a = 0.6 m/s2
Representation of the problem with the simulation.
So, we can see here that the calculated value of the acceleration was correct. The
man started from rest speeding up around 5.4 seconds until he reaches the
position 8.7m.
References
Zyl, Leandi Van. Business standard. 9 de october de 2015. february de 2022.