course PHy 201
ph1_diagnostic_quiz_0909081.� State the definition of rate of change.
ave. rate of change= change in A/ change in B
`dg1
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2.� State the definition of velocity.
average velocity= change in position/ change in clock time
`dg2
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3.� State the definition of acceleration.
average acceleration= (Vf-V0)/change in time
`dg3
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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?�
10cm/s , you obtain it by subtracting the velocity that it begins at from the velocity it ends at.
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What is its change in position and how do you obtain it from the given information?�
30cm, you subtract the position it starts at from the position it ends at.
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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,
80cm/s -30cm/s = 50cm/s , Velocity= Vf-Vo
this isn't how you find average velocity; this is change in velocity
among other things note that 50 cm/s is closer to 30 cm/s than to 80 cm/s
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and
its acceleration during this interval.�
ave.A= (80cm/s -30cm/s)/10s= 5cm/s , Acceleration= (Vf-Vo)/time
good except your units are wrong
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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.�
The rise is equal to the subtraction of the two points, 40cm/s - 10cm/s= 30cm/s, the rise is 30
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The run of the graph trapezoid.�
The base of the graph is also the run, so the run is 6
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The slope associated with the trapezoid.�
The slope is equal to the rise/run. (30cm/s)/6s= 5cm
&#The units of your final result do not follow from the units of the quantities you used to calculate it. &#
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The dimensions of the equal-area rectangle associated with the trapezoid.�
altitude= 25 on both sides, Base= 6
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The area of the trapezoid.
(25cm/s)x(6s)= 150cm/s
`dg27
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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= (Vf-Vo)m, In order to figure out the rise you need to subtract the final initial velocity from the final velocity. The rise is in meters
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What is the run of the graph trapezoid and what are its unit?�
The run is equal to the length of the base of the trapezoid in seconds. The run is in seconds
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What is the slope of the trapezoid and what are its units?�
The slope is (Vf-Vo)in m/s /(the length of the base in seconds). In order to figure out the slope I needed to first come up with a prediction of what would create the rise and the run. The run was simple because it only consists of the base of the trapezoid, which is measured in seconds. The rise of between the two points is equal to the final velocity minus the initial velocity.
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What is the area of the trapezoid and what are its units?�
The Area= [Vf-(Vf-Vo)m]*(the length of the base in seconds)
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What, if anything, does the slope represent?�
The slope represents the change in the velocity of the ball per second as it travels down the ramp.
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What is the altitude of the equal-area rectangle and what are its units?�
The altitude of the equal-area rectangle is=Vf-(Vf-Vo)m or Vo+(Vf-Vo)m
this isn't correct; for example vf - (vf - v0) = v0, and v0 +(vf - v0) = vf. Neither expression indicates the average velocity.
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What is the base of the equal-area rectangle and what are its units?�
It is the same as it was in the trapezoid. Its units are still in seconds.
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What, if anything, does the area represent?�
It represents the area of the trapezoid. If it represents anything else please explain it to me.
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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 is (Vf-Vo)(m/s) /(change in Seconds). In order to figure out the slope I needed to first come up with a prediction of what would create the rise and the run. The run was simple because it only consists of the base of the trapezoid, which is measured in seconds. The rise of between the two points is equal to the final velocity minus the initial velocity.
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What is the area of the trapezoid and what are its units?�
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What, if anything, does the slope represent?�
The slope represents the change in the velocity of the ball per second as it travels down the ramp.
That's a good answer.
You should also note that the slope represents (change in velocity) / (change in clock time), which by one definition is the average rate of change of velocity with respect to clock time, which by another definition is acceleration.
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What, if anything, does the area represent?�
I need this one explained.
The average altitude represents the average velocity and the run represents the change in clock time; to get the area you multiply these two quantities/.
What does it mean to multiply average elocity by change in clock time?
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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?� ��
It takes 3 seconds for the ball to reach 50 cm because it starts moving at a speed of 5cm/s and ends at a speed of 15cm/s this indicates that inbetween the 20 cm starting point and the 50cm ending point there is an increase in the velocity of the ball by 10cm/s. From this we can deduce the time by subtracting the starting point from the ending point and dividing that by the acceleration (10m/s). This gives us the number of seconds it takes for the ball to reach the ending point. The ball accelerates 10 meters for every second.
time= (50m-20m)/ 10m/s= 3 seconds
This isn't a bad attempt.
However the ball's average velocity is 10 cm/s (halfway between init and final vel).
Its change in position is 30 cm.
ave vel = (change in pos) / (change in clock time) so
change in clock time = (change in pos) / (average vel) = 30 cm / (10 cm/s) = 3 s.
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Good thinking and reasoning, but you need to be a little more precise about meanings, and watch your units calculations.
&#See my notes and let me know if you have questions. &#