course Phy 121
Be sure you include a copy of your access code with everything you submit. I don't always notice when you don't include it, and if that happens your work won't get posted. Even if I do notice, I have to go and look up the code and insert it. I don't mind doing that once in awhile, but if it happens consistently it begins to take time away from the more important process of responding to student work.
The last form I sent contained Week 2, Quiz 2, Version 1.This is Randomized Problems, Week 3, Quiz 1, Version 7.
Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.
DIRECT REASONING:
vo = 10m/s
aAve = 2m/s^2
`dt = 44s
`dv = (a)(`dt)= (2m/s^2)(44s) = 88m/s
vf = `dv + vo = 88m/s + 10m/s = 98m/s
`ds = (vAve)(`dt)
vAve = (vf + vo)/2 = (98m/s + 10m/s)/2 = (108m/s)/2 = 54m/s
`ds = (54m/s)(44s) = 2376 meters
FLOW CHART:
I will describe the flow of the order I did things.
First level: We were given vo, aAve, and `dt
Second level: Used a and `dt to find `dv --> (a)(`dt)= `dv
Third level: Used `dv and vo to find vf --> `dv + vo = vf
Fourth level: Used vo and vf to find vAve --> (vo + vf)/2 = vAve
Fifth level: Used vAve and `dt to find `ds --> (vAve)(`dt)= `ds
USING UNIFORM ACCELERATION EQUATIONS:
`ds = (vo)(`dt) + 1/2(a)(`dt)^2
`ds = (10m/s)(44s) + 1/2(2m/s^2)(44s)^2
`ds = 440m + 1/2(3872m) = 440m + 1936m = 2376 meters
vf^2 = vo^2 + 2(a)(`ds)
vf^2 = (10m/s)^2 + 2(2m/s^2)(2376m)
vf^2 = 100(m/s)^2 + 9504(m/s)^2 = 9602m^2/s^2
TAKE SQUARE ROOT OF BOTH SIDES
vf = 98m/s
An object has an initial velocity of 3cm/s. It travels 15cm at a uniform acceleration of 1.5cm/s^2. What is the final velocity and the time interval?
vo = 3cm/s
aAve = 1.5cm/s^2
`ds = 15cm
vf^2 = vo^2 + 2(a)(`ds)
vf^2 = (3cm/s)^2 + 2(1.5cm/s^2)(15cm)
vf^2 = 9(cm/s)^2 + 45(cm/s)^2 = 54cm^2/s^2
TAKE SQUARE ROOT OF BOTH SIDES
vf = 7.35 cm/s
vAve = (vf + vo)/2 = (7.35cm/s + 3cm/s)/2 = 10.35cm/s /2 = 5.17cm/s
`dt = `ds/vAve
`dt = 15cm/(5.17cm/s) = 2.9 seconds"
Excellent work. Let me know if you have questions.