Velocity and Acceleration - Position versus Time
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Please give the solution and a step-by-step way to figure the answers out please. Learning is important.
Use the data for all these questions on position versus time. The graph points are at 8 meters t=0 seconds. -4 meters at 1 second, -8 meters at 2 seconds, -4 meters at 3 seconds, 0 meters at 4 seconds and 28 meters at 5 seconds.
1) What is the change in position of the object during the time from 1 second to 2 seconds?
a) -8 meters
b) -4 meters
c) 0 meters
d) 4 meters
e) 8 meters
2) What is the average velocity of the object between 0 and 2 seconds.
a) -8 meters/s
b) -4 meters/s
c) 0 meters/s
d) 4 meters/s
e) 8 meters/s
3) What was the avergae velocity of the object between t=1 and t=5 seconds?
a) -4 meters/s
b) 6 meters/s
c) 0 meters/s
d) 4 meters/s
e) 8 meters/s
4) What was the acceleration of the object during the time shown?
a) -4 meters/s^2
b) 4 meters/s^2
c) -8 meters/s^2
d) 8 meters/s^2
e) 0 meters/s^2
5) What was the intial velocity of the object at t=0 seconds? Hint: calculate the velocity at different times, then make a graph of velocity versus time.
a) -16 meters/s
b) -12 meters/s
c) -4 meters/s
d) 0 meters/s
e) 4 meters/s
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Please give solution and step by step way to figure answers out please. Learning is important.
Use the data for all these questions on position versus time. The graph points are at 8 meters t=0 seconds. -4 meters at 1 second, -8 meters at 2 seconds, -4 meters at 3 seconds, 0 meters at 4 seconds and 28 meters at 5 seconds.
1) What is the change in position of the object during the time from 1 second to 2 seconds?
a) -8 meters
b) -4 meters
c) 0 meters
d) 4 meters ...
Education
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
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- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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