course Phy 201
Sept 10, 2009 8:05 p.m.
1.� State the definition of rate of change.vvvv
The average rate of change is defined by the change in position A with respect to Change in B
The definition does not involve position; you give the correct definition in the line below.
Average rate of change = Change in A / Change in B
2.� State the definition of velocity.
The average velocity of an object is defined as the change in position with respect to clock time
right idea but you don't have all the words required by the definition
Velocity = Change in position / Change in Clock time
The definition does give you this for average velocity, but this is the result of the definition, not the definition itself.
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3.� State the definition of acceleration.
The average acceleration is defined as the change in velocity with respect to clock time
Acceleration = Change in velocity / Change in Clock Time
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4.� A ball rolls along a path, moving from position 20 cm to position 50 cm as its velocity increases from 5 cm/s to 15 cm/s.
• What is its change in velocity and how do you obtain it from the given information?� &&&&
To obtain the change in velocity you subtract the final velocity from the initial velocity.
15 cm/s - 5 cm/s = 10 cm/s
• What is its change in position and how do you obtain it from the given information?� &&&&
To obtain the change in position you subtract the final position from the initial position
50 cm – 20 cm = 30 cm
5.� A ball accelerates from velocity 30 cm/s to velocity 80 cm/s during a time interval lasting 10 seconds.
Explain in detail how to use the definitions you gave above to reason out
• the average velocity of the ball during this interval, &&&&
First you take the final velocity and subtract the initial from it and it give up the velocity for the that interval.
'the velocity' is not a term that's acknowledged in this course. You have to modify the word 'velocity'.
The correct modifier for your statement is 'change in velocity', which is very different from 'the velocty' and has a very different meaning.
If you use modifiers carefully you will avoid a lot of errors.
Then you take that velocity and divide it by the clock time and that gives you the average velocity.
80 cm/s – 30 cm/s = 50 cm/s
50 cm/ s / 10 s = 5 cm/s your units aren't correct
• its acceleration during this interval.� &&&&
To find the average acceleration you take the change in velocity and divide it by the clock time.
5 cm/s / 10 s = 0.5 cm/ s
To find the average acceleration you take the change in velocity and divide it by the change in clock time.
5 cm/s is not the change in velocity.
Remember, the main goal is to use a detailed reasoning process which connects the given information to the two requested results.� You should use units with every quantity that has units, units should be included at every step of the calculation, and the algebraic details of the units calculations should be explained.
6. �A �graph trapezoid� has �graph altitudes� of 40 cm/s and 10 cm/s, and its base is 6 seconds.� Explain in detail how to find each of the following:
• The rise of the graph trapezoid.� &&&&
Rise = Vf – Vo = 40 cm/s -10 cm/s = 30 cm/s
• The run of the graph trapezoid.� &&&&
Run = dt = 6 seconds
• The slope associated with the trapezoid.� &&&&
Slope = Rise / Run = 30 cm/s / 6 seconds = 5 cm/s
the units of your final result aren't right
• The dimensions of the equal-area rectangle associated with the trapezoid.� &&&&
The altitude of the rectangle is 40 cm/s + 10 cm/s divided by 2 which gives you 25 cm/s for the sides of the equal area triangle
• The area of the trapezoid. &&&&
The area of the equal area triangle is the altitude multiplied by the base.
25 cm/s * 6 seconds = 150 cm
Each calculation should include the units at every step, and the algebraic details of the units calculations should be explained.
7.� If the altitudes of a �graph trapezoid� represent the initial and final positions of a ball rolling down an incline, in meters, and the based of the trapezoid represents the time interval between these positions in seconds, then
• What is the rise of the graph trapezoid and what are its units?� &&&&
The rise equals Vf – Vo and the units are in meters/seconds
• What is the run of the graph trapezoid and what are its unit?� &&&&
The run equals the base of the trapezoid or dt and the units are in seconds
• What is the slope of the trapezoid and what are its units?� &&&&
The slope of the trapezoid is equal to Vf – Vo / dt and the units are meters/seconds
those aren't the units
• What is the area of the trapezoid and what are its units?� &&&&
The area of the trapezoid is equal to both of the altitudes / 2 * the base of the trapezoid and the units are in meters
• What, if anything, does the slope represent?� &&&&
The change in velocity
The change in velocity is vf - v0. The slope is (vf - v0) / `dt.
• What is the altitude of the equal-area rectangle and what are its units?� &&&&
The altitude of the equal area rectangle is Vf + Vo / 2, the units are in meters/seconds
• What is the base of the equal-area rectangle and what are its units?� &&&&
The base of the rectangle is equal to dt and the units are in seconds
• What, if anything, does the area represent?� &&&&
The area represents the change in position
Each answer should include a complete explanation, reasoned out from the geometry of the trapezoid and the definitions you gave at the beginning.
8.� If the altitudes of a �graph trapezoid� represent the initial and final velocities of a ball rolling down an incline, in meters / second, and the based of the trapezoid represents the time interval between these velocities in seconds, then
• What is the slope of the trapezoid and what are its units?� &&&&
The slope of the trapezoid equals the final velocity minus the initial velocity divided by base or delta t. And the units are in meters/ seconds
the slope is a rate of change, and its units are not the same as those of change in velocity
• What is the area of the trapezoid and what are its units?� &&&&
The area of the trapezoid equals the final velocity plus the initial velocity divided by 2 which gives you the altitude and then you multiply the altitude by the base and that gives you the area of the trapezoid and the units are in meters.
• What, if anything, does the slope represent?� &&&&
Change in velocity
see previous notes
• What, if anything, does the area represent?� &&&&
Change in position
Each answer should include a complete explanation, reasoned out from the geometry of the trapezoid and the definitions you gave at the beginning.
9.� A ball rolls along a path, moving from position 20 cm to position 50 cm as its velocity increases from 5 cm/s to 15 cm/s.
• If its acceleration is uniform, then how long does this take, and what is the ball�s acceleration?� ��&&&&
50 cm – 20 cm = 30 cm
15 cm/s – 5 cm/s = 10 cm/s 30 cm / 10 cm/s = 3 seconds
15 cm/s – 5 cm/s = 10 cm/s
Change in acceleration = 10 cm/s / 3 seconds = 3.3333333 cm/s
"
I'm assessing your work on several criteria, each assessed on a 5-point scale. This isn't a grade and won't be used as a grade. It's for your information.
On the 5-point scale, 1 represents something you seldom did well and 5 something you almost always did well. On the average, most of the class at this point appears to be around the middle of the scale.
Here are the criteria, listed more or less in the order of importance at this stage of the course:
Overall level of reasoning process (i.e., are you basing your answers on the definitions and explaining your details).
My assessment on this criterion: 4.
accuracy of definitions: 2 (could easily be improved to a 5)
accuracy of wording: 4
consistent use of units: 4
correct identification of quantities (still a weak area for the class as a whole): 3+
correct units calculations (still a weak area for the class as a whole): 2
I recommend that you submit a revision of your work, according to my notes. You doing well on most of these criteria, and the ones you've overlooked should be easy to fix, so try to 'lock in' all of these ideas as soon as possible.