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PHY-150 Intro Physics: Module Two Assignment
C Daniel T Bundor
Southern New Hampshire University
PHY-150-H7663 Intro Physics: Mechanics 24EW4
Professor Ralph Spraker
March 30, 2024
PHY-150 Intro Physics: Module Two Assignment
Activity 1: Graph and interpret motion data of a moving object.
Activity 1. Table 1
Time (x axis) (seconds)
Position (y axis)
(meters)
0
0
5
20
10
40
15
50
20
55
30
60
35
70
40
70
45
70
50
55
Questions for Activity 1
Question 1:
What is the average speed of the train during the time interval from 0 s to 10 s?
Average Speed = Total Distance / Total Time = 40 meters / 10 seconds
= 4 meters per second
Question 2:
Using the equation: v
=
s
2
−
s
1
t
2
−
t
1
, calculate the average speed of the train as it moves from position x = 50m to x = 60m.
V=60m-50m
30s-15s
V=10m
15s
V=10m
15s
V=0.667 m/s
Question 3:
What does the slope of the line during each time interval represent?
The slope of the line on a position-time graph represents the velocity of the object being tracked.
Question 4:
From time t = 35 s until t = 45 s, the train is located at the same position. What is slope of the line while the train is stationary?
The slope of the line while the train is stationary (from t=35t=35 s to t=45t=45 s) is zero. This is because the position of the train remains constant during this time interval, meaning there is no change in position with respect to time.
Question 5:
Calculate the average speed of the train as it moves from position x = 70m to x = 55m. What does the sign of the average velocity during this time interval represent? Total Time=t2−t1=50s−40s=10s
Now, plug these values into the formula for average speed:
Average Speed=15 m10 s=1.5 m/s
The sign of the average velocity during this time interval represents the direction of motion. Since the velocity is positive (1.5 m/s), it indicates that the train is moving towards the positive direction, which means it's moving to the left on the position-
time graph.
Question 6:
What is the displacement of the train from time t = 0s until t = 50s?
Δx=x2−x1
Δx=55 m−0
Δx=55
m
So, the displacement of the train from t=0s until t=50s is 55 meters.
Question 7:
What is the total distance traveled by the train from time t = 0s until t = 50s?
Total Distance=
∣
20−0
∣
+
∣
40−20
∣
+
∣
50−40
∣
+
∣
55−50
∣
+
∣
60−55
∣
+
∣
70−60
∣
+
∣
70−70
∣
+
∣
70−
70
∣
+
∣
55−70
∣
Total Distance=20+20+10+5+5+10+0+0+15
Total Distance=85 meters So, the total distance traveled by the train from t=0t=0 s until t=50t=50 s is 8585 meters.
Question 8. What is the slope of the line during the time interval t = 45 to t = 50?
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