may heip you calculate the area). Insert below a screen shot of the simulation for above description and your worksheet which shows how you obtain final answer for each question Chats 39 seconds Velocity 120- a. What was the distance covered? m b. What is the area under the velocity curve? m c. What was the displacement of the man? m d. What was the speed of the man? (Remember speced distance traveled/given time interval) m/s e. What was the velocity? (Remember velocity = displacement/given time interval) m/s and direction (if any)

Inquiry into Physics
8th Edition
ISBN:9781337515863
Author:Ostdiek
Publisher:Ostdiek
Chapter1: The Study Of Motion
Section: Chapter Questions
Problem 9C
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12. The area under the velocity vs time graph is supposed to be equal to the distance traveled along a straight line.
Displacement final position - initial position
Distance: the distance traveled along the actual path.
Since the average velocity = displacement (distance traveled along a straight line) / given time interval
1/p = a
d = vt
Set the position of the man at -10.0 m, the velocity equal to 5 m/s and the acceleration to 0.0. Run the simulation until the Moving Man hits the wall. What is the area under the velocity curve when the man reaches the +10 m mark? (You want to calculate the area of the shape between the
velocity curve and the horizontal time axis. Realizing what kind of shape it is may help you calculate the area).
Insert below a screen shot of the simulation for above description and your worksheet
which shows how you obtain final answer for each question
120
a. What was the distance covered?
m
b. What is the area under the velocity curve?
c. What was the displacement of the man?
d. What was the speed of the man?
(Remember speed = distance traveled/given time interval)
m/s
e. What was the velocity?
(Remember velocity = displacement/given time interval)
m/s and direction (if any)
Transcribed Image Text:12. The area under the velocity vs time graph is supposed to be equal to the distance traveled along a straight line. Displacement final position - initial position Distance: the distance traveled along the actual path. Since the average velocity = displacement (distance traveled along a straight line) / given time interval 1/p = a d = vt Set the position of the man at -10.0 m, the velocity equal to 5 m/s and the acceleration to 0.0. Run the simulation until the Moving Man hits the wall. What is the area under the velocity curve when the man reaches the +10 m mark? (You want to calculate the area of the shape between the velocity curve and the horizontal time axis. Realizing what kind of shape it is may help you calculate the area). Insert below a screen shot of the simulation for above description and your worksheet which shows how you obtain final answer for each question 120 a. What was the distance covered? m b. What is the area under the velocity curve? c. What was the displacement of the man? d. What was the speed of the man? (Remember speed = distance traveled/given time interval) m/s e. What was the velocity? (Remember velocity = displacement/given time interval) m/s and direction (if any)
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