course phy 201
September 17 at 1049 am
1. State the definition of rate of change.vvvv
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Rate of change is the change in quantity A with respect to quantity B.
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2. State the definition of velocity.
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Velocity is the rate of change of position with respect to clock time.
3. State the definition of acceleration.
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Acceleration is the average 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 10 cm/s. You can get this change in velocity by taking the final velocity (15cm/s) and subtracting the initial velocity (5 cm/s)
• What is its change in position and how do you obtain it from the given information? &&&&
The change in position is 30 cm. You can get this change in position by taking the final position (50 cm) and subtract the initial position (20 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, &&&&
To find the average velocity of a ball that starts out at 30 cm/ sec and goes to 80 cm./ sec you would first need to add the two velocities together and get a total of 110 cm/s. Then you would take that 110 cm/s and divide it by two to get the average of the velocities. This average velocity would be 55 cm/s.
and
• its acceleration during this interval. &&&&
To find the average accceleration you first need to find the change in velocity. To find this we must take the final velocity (80cm/s) and subtract the initial velocity (30 cm/s). This gives you a difference of 50 cm/s. Then we get the change in clock time which is given in the question (10 seconds) and divide the 50 cm/s by the 10 second interval. The answer turns out to be 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. &&&&
The rise of the graph is 30 cm/s. You can get this by finding the difference in the two altitude heights.
• The run of the graph trapezoid. &&&&
The run of the trapezoid is the same as the base of the trapezoid. So the run of the graph is 6 seconds.
• The slope associated with the trapezoid. &&&&
The slope of the graph comes from dividing the rise of the graph by the run of the graph. In this graph the rise is 30 cm/s and the run is 6 seconds. So it you were to divide 30 cm/s by 6 seconds you would get 5 cm/ sec^2
• The dimensions of the equal-area rectangle associated with the trapezoid. &&&&
• To find the dimensions of the equal area rectangle you would first find the average height of the two altitudes. This is done by adding the two altitudes together and then dividing that quantity by two. In the case of this particular trapezoid the average of the two altitudes is 25 cm/s. The other dimension of the equal area rectangle is equal to the base of the trapezoid (6 seconds)
• The area of the trapezoid. &&&&
The area of the trapezoid would be caculated using the demensions of the equal area rectangle that can be constructed from the trapezoid. These demensions are 25 cm/s by 6 seconds. To find the area you simply multiply these two quantites together to get 150 cm which is the area of the trapezoid as well as the equal area rectangle.
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 difference between the final position and the initial position. In other words the rise is the change in position. The unit of measure of the rise of the graph is meters.
• What is the run of the graph trapezoid and what are its unit? &&&&
The run of the graph represents the change in clock time over which the change in position took place. The run will equal the measurement of the base of the trapezoid which in this case the unit used is seconds.
• What is the slope of the trapezoid and what are its units? &&&&
The slope of the trapezoid is the change in position with respect to clock time which is velocity. The units for this would be meters/second.
• What is the area of the trapezoid and what are its units? &&&&
The are of the trapezoid is calculated by taking the average of the lengths of the altitudes and multiplying it by the base of the trapezoid. The units of measure of the area of the trapezoid would be meters*seconds. This comes from simply multiplying together the units if the altitudes (meters) and the unit of the base of the trapezoid (seconds)
• What, if anything, does the slope represent? &&&&
The slope of this trapezoid represents the velocity.
• What is the altitude of the equal-area rectangle and what are its units? &&&&
The altitude of the equal area rectangle would be calculated by adding the two altitudes together and dividing by 2. The unit of measure of the altitude would be meters.
• What is the base of the equal-area rectangle and what are its units? &&&&
The base of the equal area rectangle would equal the measurment of the base of the original trapezoid. The unit of measure of the base of the equal area rectangle would be seconds.
• What, if anything, does the area represent? &&&&
The area would represent the total change in position over the time interval in question.
The rise represents the change in position.
The area would be in units of m * s and would not have a meaning associated with the motion of the object.
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 is the difference in the final minus the initial velocities of the ball, divided by the base of the trapezoid. The units for this slope would be meters/second^2
• What is the area of the trapezoid and what are its units? &&&&
The area of the trapezoid would be the average of the lenghts of the two altitudes (in meters/ second) multiplied by the base of the trapezoid (in seconds). The unit of measure of the area would be meters because meters/second times seconds would give you seconds on the top and bottom of the fraction and since anything over itself is 1 you can simplify the fraction to meters*1 or just meters.
• What, if anything, does the slope represent? &&&&
• The slope for this trapezoid is the change in velocity with respect to clock time which means the slope of the trapezoid equals the acceleration.
• What, if anything, does the area represent? &&&&
The area of this trapezoid represents the total distance traveled over the 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? &&&&
In order to get the value the change in clock time or ‘dt you need to plug what you know into the equation ‘ds= (vf+v0/2)* ‘dt. This gives you 30 cm= ((5cm/s+ 15 cm/s)/2) * ‘dt. First add the 5 cm/s and the 15 cm/s together to get 30 cm= (20 cm/s / 2)* ‘dt. Then divide the 20 cm/s by 2 to get 30 cm= 10cm/s * ‘dt. Then divide both sides by 10 cm/s to get the value for ‘dt which is 3 seconds.
In order to get the value for acceleration (a) you plug what you know into the equation a=(vf-v0)/ ‘dt which gives you a= (15cm/s- 5 cm/s)/ 3 seconds. First subtract the 5 cm/s from the 15 cm/s to get a= 10 cm/s/ 3 seconds. Then divide the 10 cm/s by 3 seconds to get the acceleration which is 3.33333 cm/s^2
Excellent work. It appears that you've mastered these ideas.
See my one note.