course PHY 121
Trying really hard to catch on will work on lab tomorrow
006. `query 6*********************************************
Question: Given uniform acceleration 2 m/s^2, displacement 125 meters and final velocity 30 m/s, a student finds that the initial velocity is zero and the time interval is 10 seconds. We wish to check to see if the student's results are consistent with the given information.
• Using the initial velocity and time interval obtained by the student, along with the 30 m/s final velocity, quickly reason out the acceleration and the displacement in terms of the definitions of average velocity and/or average acceleration and the assumption of uniform acceleration.
• State whether the student's solutions are consistent with the originally given information.
• Compare with your solution to this problem.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
#$&* a) the average velocity is 15 m/s, which since acceleration is uniform is simply the average of the initial and final velocities. multiplying the 15 m/s average velocity by the 10 s time interval interval we get displacement 150 m. the change in velocity would be 30 m/s, since the velocity changes on this interval from 0 m/s to 30 m/s; so the acceleration would be aAve = `dv / `dt = 30 m/s / (10 s) = 3 m/s^2.
b) The student's solution is not consistent with the given information, which specified acceleration 2 m/s^2 and displacement 125 meters.
c) v0 = +- sqrt( vf^2 - 2 a `ds) =
+- sqrt( (30 m/s)^2 - 2 * (2 m/s^2) * (125 m) ) =
+- sqrt( 900 m^2 / s^2 - 500 m^2 / s^2) =
+- sqrt( 400 m^2 / s^2) =
+- 20 m/s.
vf = 20 m/s
ds = (vf + v0) / 2 * `dt gives us `dt = 2 * `ds / (vf + v0) = 2 * 125 m / (20 m/s + 30 m/s) = 5 s.
vf = -20 m/s
`ds = (vf + v0) / 2 * `dt gives us `dt = 2 * `ds / (vf + v0) = 2 * 125 m / (-20 m/s + 30 m/s) = 25 s.
confidence rating #$&* ok
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): ok
Self-critique rating #$&* ok
*********************************************
Question:
An automobile coasts at high speed down a hill, speeding up as it goes. It encounters significant air resistance, which causes it to speed up less quickly than it would in the absence of air resistance.
If the positive direction is down the hill, then
• Is the direction of the automobile's velocity positive or negative?
• Is the direction of the air resistance positive or negative?
If the positive direction is up the hill, then
• Is the direction of the automobile's velocity positive or negative?
• Is the direction of the automobile's acceleration positive or negative?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
#$&* If the direction down the hill is positive then the velocity is positive, the acceleration is positive and air resistance is negative. If the direction down the hill is negative then the velocity is negative, the acceleration is negative and air resistance is positive.
confidence rating #$&* ok
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): ok
------------------------------------------------
Self-critique rating #$&* ok
"
&#Good work. Let me know if you have questions. &#
#$&*
course PHY 121
Trying really hard to catch on will work on lab tomorrow
006. `query 6*********************************************
Question: Given uniform acceleration 2 m/s^2, displacement 125 meters and final velocity 30 m/s, a student finds that the initial velocity is zero and the time interval is 10 seconds. We wish to check to see if the student's results are consistent with the given information.
• Using the initial velocity and time interval obtained by the student, along with the 30 m/s final velocity, quickly reason out the acceleration and the displacement in terms of the definitions of average velocity and/or average acceleration and the assumption of uniform acceleration.
• State whether the student's solutions are consistent with the originally given information.
• Compare with your solution to this problem.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
#$&* a) the average velocity is 15 m/s, which since acceleration is uniform is simply the average of the initial and final velocities. multiplying the 15 m/s average velocity by the 10 s time interval interval we get displacement 150 m. the change in velocity would be 30 m/s, since the velocity changes on this interval from 0 m/s to 30 m/s; so the acceleration would be aAve = `dv / `dt = 30 m/s / (10 s) = 3 m/s^2.
b) The student's solution is not consistent with the given information, which specified acceleration 2 m/s^2 and displacement 125 meters.
c) v0 = +- sqrt( vf^2 - 2 a `ds) =
+- sqrt( (30 m/s)^2 - 2 * (2 m/s^2) * (125 m) ) =
+- sqrt( 900 m^2 / s^2 - 500 m^2 / s^2) =
+- sqrt( 400 m^2 / s^2) =
+- 20 m/s.
vf = 20 m/s
ds = (vf + v0) / 2 * `dt gives us `dt = 2 * `ds / (vf + v0) = 2 * 125 m / (20 m/s + 30 m/s) = 5 s.
vf = -20 m/s
`ds = (vf + v0) / 2 * `dt gives us `dt = 2 * `ds / (vf + v0) = 2 * 125 m / (-20 m/s + 30 m/s) = 25 s.
confidence rating #$&* ok
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): ok
Self-critique rating #$&* ok
*********************************************
Question:
An automobile coasts at high speed down a hill, speeding up as it goes. It encounters significant air resistance, which causes it to speed up less quickly than it would in the absence of air resistance.
If the positive direction is down the hill, then
• Is the direction of the automobile's velocity positive or negative?
• Is the direction of the air resistance positive or negative?
If the positive direction is up the hill, then
• Is the direction of the automobile's velocity positive or negative?
• Is the direction of the automobile's acceleration positive or negative?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
#$&* If the direction down the hill is positive then the velocity is positive, the acceleration is positive and air resistance is negative. If the direction down the hill is negative then the velocity is negative, the acceleration is negative and air resistance is positive.
confidence rating #$&* ok
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): ok
------------------------------------------------
Self-critique rating #$&* ok
"
&#Your work looks good. Let me know if you have any questions. &#
#$&*