#$&*
phy121
Your 'cq_1_03.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_03.1_labelMessages.txt **
The problem:
A ball starts with velocity 0 and accelerates uniformly down a ramp of length 30 cm, covering the distance in 5 seconds.
• What is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve=’ds/’dt
vAve= 30cm/5s
vAve= 6cm/s
#$&*
• If the acceleration of the ball is uniform then its average velocity is equal to the average of its initial and final velocities.
You know its average velocity, and you know the initial velocity is zero.
What therefore must be the final velocity?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve= (initial velocity + final velocity)/2
6cm/s= (0 + final)/2
6cm/s*2= 0 + final
12cm/s= the final velocity.
#$&*
• By how much did its velocity therefore change?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The velocity changes by 12cm/s from rest until the end. Its final velocity is 6cm/s more than the average velocity.
#$&*
• At what average rate did its velocity change with respect to clock time?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Velocity/time
(12cm/s)/5s
2.4cm/s
#$&*
• What would a graph of its velocity vs. clock time look like? Give the best description you can.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
since the object started from rest at 0 seconds, that’s where the first point would be. We know by the end of the roll, the ball was travelling at 12cm/s at 5 seconds. We can see that the graph has the time on the x-axis and the velocity on the y-axis. The graph increases at an increasing rate. Since the ball is accelerating at a uniform speed, that means it’s a constant movement, which would make the graph have a straight line increasing at all times.
#$&*
** **
15 minutes
** **
self-critique #$&*
#$&* self-critique
self-critique rating
`gr91 rating #$&*:
`c031
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
________________________________________
`gr91
#$&*
#$&*
phy121
Your 'cq_1_03.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_03.1_labelMessages.txt **
The problem:
A ball starts with velocity 0 and accelerates uniformly down a ramp of length 30 cm, covering the distance in 5 seconds.
• What is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve=’ds/’dt
vAve= 30cm/5s
vAve= 6cm/s
#$&*
• If the acceleration of the ball is uniform then its average velocity is equal to the average of its initial and final velocities.
You know its average velocity, and you know the initial velocity is zero.
What therefore must be the final velocity?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve= (initial velocity + final velocity)/2
6cm/s= (0 + final)/2
6cm/s*2= 0 + final
12cm/s= the final velocity.
#$&*
• By how much did its velocity therefore change?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The velocity changes by 12cm/s from rest until the end. Its final velocity is 6cm/s more than the average velocity.
#$&*
@&
The ball didn't begin the interval with its average velocity.
*@
• At what average rate did its velocity change with respect to clock time?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Velocity/time
(12cm/s)/5s
2.4cm/s
@&
This is the right calculation, but it's not velocity / time. By the definition, it's change in velocity / change in clock time.
Velocity / time would work if initial velocity and initial time were both zero, but you don't want to even think velocity / time.
Note that the units are not cm/s. You're dividing cm/s by s.
*@
#$&*
• What would a graph of its velocity vs. clock time look like? Give the best description you can.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
since the object started from rest at 0 seconds, that’s where the first point would be. We know by the end of the roll, the ball was travelling at 12cm/s at 5 seconds. We can see that the graph has the time on the x-axis and the velocity on the y-axis. The graph increases at an increasing rate. Since the ball is accelerating at a uniform speed, that means it’s a constant movement, which would make the graph have a straight line increasing at all times.
#$&*
** **
15 minutes
** **
@&
You're on the right track, but you're making some errors.
Check out my notes and let me know if you have questions.
No need to resubmit unless you want me to take another look.
*@