course 201
9/10 11AM
1. State the definition of rate of change.vvvv
The average rate of change of A with respect to B is the change in A divided by the change in B.
2. State the definition of velocity.
Velocity is the rate of change of position with respect to clock time.
3. State the definition of acceleration.
Acceleration is the rate of change of velocity with respect to clock time.
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?
• The change in velocity is final velocity – 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?
• The change in position is final position- 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
• The average velocity will be the average of the initial and final velocities, or (Vf+V0)/2= 55 cm/s. This corresponds to the new altitude of the equal area rectangle, if we use the given data to depict a trapezoid.
and
• its acceleration during this interval.
• The acceleration is found by using the formula `dvelocity/`dclocktime, which on the illustration of a graph trapezoid using this data corresponds with the slope. (Vf-V0)/`dt=(80-30)/10=5 cm sec^2
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 is the change in y= 40-10=30
• The run of the graph trapezoid.
• Run is the change in x= 6
• The slope associated with the trapezoid.
• Slope=rise/run=30/6=5
• The dimensions of the equal-area rectangle associated with the trapezoid.
• The base stays the same, but the new altitude is found by averaging the two given altitudes. (Vf+V0)/2=(40+10)/2=25, so the dimensions of the equal area rectangle are 6 by 25
• The area of the trapezoid.
• The area of the trapezoid is easier to find using the area of the equal area rectangle, which I found to have an altitude of 25 cm/s and a base of 6 cm. Area=base*height=6 sec*25 cm/s =150 cm, meaning there has been a change in position over this time interval of 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 is the change in y, which would be the difference between the final and initial positions of the ball, given in meters
• What is the run of the graph trapezoid and what are its unit?
• The run is the change in x, which would be the change in clock time, given in seconds
• What is the slope of the trapezoid and what are its units?
• The slope of the trapezoid is rise/run, or `d position/`d time, which gives units of meters/second, which corresponds with velocity
• What is the area of the trapezoid and what are its units?
• The area will use the average of the initial and final positions in order to give an altitude to the equal area rectangle. the same base in seconds will be used. Area will then be `dtime*(Vf+V0)/2, which will form units of meter seconds (????)
• What, if anything, does the slope represent?
• The slope represents the velocity, as rise/run corresponds to `dposition/`dclocktime, the definition of velocity
• What is the altitude of the equal-area rectangle and what are its units?
• The altitude of the equal area rectangle is the average of the initial and final positions, (Vf+V0)/2. It’s units will remain in meters.
• What is the base of the equal-area rectangle and what are its units?
• The base of the equal area rectangle remains the same as the trapezoid, `dtime with its units in seconds
• What, if anything, does the area represent?
• The unit of meter seconds is new to me, I’m not sure what it implies
Good answer. That quantity is pretty much meaningless for the analysis of motion.
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 would be rise/run=`d velocity/`dtime= (Vf-V0)/`dtime, which will arrive at units of meters/second^2, indicating acceleration
• What is the area of the trapezoid and what are its units?
• The area of the trapezoid can be found using the base given, and a new altitude can be calculated by finding the average of the two given altitudes. Area will be `dt*{(Vf+V0)/2}, and its units will be in meters, implying that this value represents the change in position that occurred over this time interval
• What, if anything, does the slope represent?
• The slope represents the average rate of acceleration
• What, if anything, does the area represent?
• The area represents the change in position that occurred over this time interval
• 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?
• Since we know that change in position can be calculated by finding the area of the graph trapezoid, and we are given the change in position, we can solve for the unknown base value. A=base*height, `dposition=`dtime*{(Vf+V0)/2}, 50 cm-20 cm=`dtime*{(15cm/s+5cm/s)/2}, 30 cm=`dtime*10cm/s, `dtime=3 seconds
• To find the acceleration, we need to calculate the slope of the trapezoid, which we can now do since we have the base value. Slope=rise/run=change in y/change in x=(Vf-V0)/`dt=(15 cm/s-5cm/s)/3 seconds=10cm/s /3 sec=3.33 cm/sec^2
Excellent work. I can't find a flaw. You're doing top-notch work.
I'm giving most students detailed ratings on a number of criteria, for feedback only. I'm not going to break yours down, because on a 1-5 scale, with the class as a whole at about 3 on the scale, you're at 5+ on every one.