course Phy 201
This is the homework that i have already done but i felt like that i did not understand it correctly so i will resubmit it.
The remaining problems will be assigned again. If you got them right, or very nearly right, you don't need to submit them again. If you had errors, you can submit them by simply submitting a copy of your original work, with whatever changes you deem appropriate.xxxx
On a graph of income stream in dollars per month vs. clock time in months, we find the two points (16, 1000) and (40, 1200).
What do the altitudes of the graph represent? The altitudes of this graph represent
income stream in dollars per month.
What is the rise between the two points of this graph? The 'rise' of the graph is the change in the graph altitude 1000 to 1200 so therefore the rise 200.
What is the run between these points? The 'run' of the graph is the change in the horizontal coordinate 16 to 40 so therefore the rise is 24.
What therefore is the slope associated with this graph trapezoid? slope = rise / run = (change in speed) / (change in clock time) =
What does the slope mean? slope = rise / run = (change in speed) / (change in clock time), which is the definition of the average rate of change speed with respect to clock time
What does the base of the graph represent? The base of this graph represent clock time in month.
What are the dimensions of the equal-area graph rectangle?
What is the area of the graph? Multiplying the base by the altitude we have 16*1000=16000, 40*1200=48000, 16000+48000=64000
What does the area of the graph represent? The area represents dollars per month and clock time in months. Then we multiply the income by clock time.
On a graph of force in pounds vs. position in feet, we find a graph rectangle with base 200 and altitude 30.
What do the altitudes of the graph represent? The altitudes of this graph represent force in pounds.
What is the rise between the two points of this graph? The graph would not have a rise because its only one point.
What is the run between these points? The graph would not a run because it only list one point.
What therefore is the slope associated with this graph trapezoid? slope = rise / run = (change in speed) / (change in clock time) =
What does the slope mean? slope = rise / run = (change in speed) / (change in clock time), which is the definition of the average rate of change speed with respect to clock time.
What does the base of the graph represent? The base of this graph represent position in feet.
What are the dimensions of the equal-area graph rectangle?
What is the area of the graph? Multiplying the base by the altitude we have 200*30=6000
What does the area of the graph represent? The area represents force in pounds and position in feet. Then we multiply the force by position.
On a graph of density in grams / centimeter vs. position in centimeters, we find the points (5, 12) and (20, 10).
What do the altitudes of the graph represent? The altitudes of this graph density in grams/centimeter.
What is the rise between the two points of this graph? The 'rise' of the graph is the change in the graph altitude -2.
What is the run between these points? The 'run' of the graph is the change in the horizontal coordinate 5-20 therefore the rise would be 15.
What therefore is the slope associated with this graph trapezoid? slope = rise / run = (change in speed) / (change in clock time) =
What does the slope mean? slope = rise / run = (change in speed) / (change in clock time), which is the definition of the average rate of change speed with respect to clock time.
What does the base of the graph represent? The base of this graph position in centimeters.
What are the dimensions of the equal-area graph rectangle?
What is the area of the graph? Multiplying the base by the altitude we have 5*12=60, 20*10=200, 60+200=260
What does the area of the graph represent? The area represents the density by position."
You are following the required process in analyzing and answering these problems. You are in good shape at this point, and making a very good effort which I'm confident will result in success in this course.
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course Phy 201
Galileo Experimentxxxx
Use the fastest pendulum you can reliably count.
Count your pendulum for a minute.
The instructor will operate the experiment. You just watch and time.
Time the ball as it moves down 10 feet of ramp, using a pendulum.
Repeat for 9 feet, then for 6 feet, then for 4 feet.
Graph position vs. clock time for the ball rolling down the ramp, with clock time in units of half-cycles.
Sketch the graph pretty carefully and sketch a smooth curve that you believe represents the actual position vs. clock time behavior of the ball.
The speed of the ball is constantly changing, so your graph will not contain any straight lines.
There is some uncertainty in your timing, so while your curve will probably come close to your data points, it shouldn't be expected to actually pass through any of them.
In other words, your curve should represent the actual behavior of the ball, as best you can infer it from the data points, but you shouldn't go out of your way to make your curve actually go through any of the points.
Based on your graph estimate the following:
The time required to travel down the first one-foot ramp. Due to the graph that I made earlier it looks like it would be around .5
The distance that would be traveled during each half-cycle of your pendulum (i.e., the distance from the start to the end of the first half-cycle, the distance from start to end of the second half-cycle, etc.). According to my graph and estimate the distance that would be traveled during each half cycle would be around 6 inches.
The average rate of change of position with respect to clock time on the first, the third, the fifth, the seventh and the ninth 1-foot ramps. The first would be .5/1=.5
The third would be 2.5/3=.83, The fifth would be 3.5/5.7, The ninth would be 5.0/9=.55
It's not completely clear how you got those estimates, but we'll be going over this in some detail tomorrow in class.
Trapezoids
Sketch a y vs. x coordinate system. The y axis is vertical (up and down the page), the x axis horizontal (left and right across the page).
You have a 'graph trapezoid' on your desk.
A 'graph trapezoid' has the property that one of its sides is perpendicular to two other sides.
Orient your trapezoid so that its base rests somewhere on the x axis. (The base is the side which is perpendicular to two other sides; not every trapezoid has a base in this sense, but a 'graph trapezoid' does).
Estimate the two 'graph altitudes' of your trapezoid, its 'altitude' and its 'base'. You can use any unit with which you are comfortable to make your estimate (e.g., centimeters, inches, feet, kilometers, nanometers, pounds, liters, gallons, kilograms, slugs, cubic feet, miles per hour; whatever you think works best for you is fine, though length units are probably most appropriate to this exercise). (The 'graph altitudes' are the sides which are parallel to the vertical axis when the base rests on the horizontal axis).
Make on fold in the trapezoid so that if you tear the paper along the fold, the two pieces can be reassembled to make a rectangle.
Answer the two questions below, and in your answers explain your reasoning by giving the estimated dimensions, and a complete description of what you did. Your explanation show how you proceeded from your estimates to your results.
What is the 'graph slope' associated with your trapezoid? 9.6/5=1.92
What would be the dimensions of this rectangle? 9.5 by 5
I can't tell if these results follow from your estimates because you didn't include your estimates of the sides (my fault for not asking).
Definitions of average velocity and average acceleration:
These are the central definitions for the first part of your course.
� Everything you do in analyzing motion should come back to these definitions:
� The average velocity of an object on an interval is its average rate of change of position with respect to clock time on that interval.
� The average acceleration of an object on an interval is its average rate of change of position with respect to clock time on that interval.
Analyzing the motion of the Lego racer:
We estimated that the Lego racer traveled 60 cm in 1.5 seconds to rest as it traveled in the direction opposite our chosen positive direction, then 30 cm in 1.2 seconds to rest as it traveled in our chosen positive direction.
Applying the definition of average velocity to the second motion:
By the definition, we are finding average rate of change of position with respect to clock time.
The A quantity is the position of the racer.
The B quantity is the clock time.
The average rate is by definition of average rate equal to (change in A) / (change in B).
Having identified the A and B quantities we find that
average velocity = average rate of change of position with respect to clock time = (change in position) / (change in clock time).
According to our information, the change in position is +30 cm and the change in clock time is +1.2 seconds.
Thus our average velocity is
average velocity = average rate of change of position with respect to clock time = (change in position) / (change in clock time) = (+ 30 cm) / (+1.2 s) = +25 cm / sec.
Find the average velocity for the first motion, using similar steps to connect your result with the definition of average velocity.
If we sketch a graph of velocity vs. clock time for the second motion:
we know that the velocity ended up at zero
we know that the cart was moving in the positive direction as it slowed to rest
if we assume that the graph is a straight line, we conclude that the line decreases toward a point on the horizontal axis during the 1.2 second interval (you should have a sketch of the graph in your notes)
we know that the average velocity is 25 cm / s; since the final velocity is zero we conclude that the initial velocity is greater than 25 cm / s; and since we expect the average velocity to occur at the middle of the time interval we conclude that the initial velocity was 50 cm/s
Our graph therefore forms a trapezoid with base 1.2 seconds, and altitudes 50 cm/s and 0 cm/s (in this case the trapezoid is in fact a triangle). We could find the trapezoid's associated slope and area.
The slope is rise / run. The rise is the change in velocity. Velocity changes from 50 cm/s to 0, so the change in velocity is
change in velocity = final velocity - initial velocity = 0 cm/s - 50 cm/s = - 50 cm/s.
The run is 1.2 seconds. So the slope is
slope = rise / run = - 50 cm/s / (1.2 s) = -42 (cm / s) / (s) = -42 (cm / s) * ( 1 / s) = -42 cm / s^2.
Since the rise represents change in velocity and the run represents change in clock time, our calculation gives us (change in velocity) / (change in clock time).
This is the form of an average rate of change. Recalling the definition of average rate of change, we see that velocity is the A quantity, clock time the B quantity, so that this is the average rate of change of velocity with respect to clock time.
This is the definition of acceleration.
The slope of this graph represents the acceleration of the car.
Note that our reasoning requires that the v vs. t graph be a straight line. Otherwise we could not have concluded that the initial velocity is 50 cm/s.
What is the slope of the graph of the first motion (the distance was 30 cm and required 1.5 seconds)? Average Velocity=average velocity wrt clock time; which would be 30/1.2=25cm/s
Answer by duplicating the reasoning used above.
What is the area of the graph trapezoid corresponding to the first motion, and what does this area represent? The are of the rectangle would be length* width (9.5*5=47.5), the area of the small triangle would be length * with (4*1.5=6) so the area of the trapezoid would be 47.5+6=53.5, This are represents the dimensions in the space.
Answer by identifying all the quantities you use to find the area, and as best you can reason out the meaning of your result. Reason in detail similar to that used above, though the reasoning process will be different for the question of area.
The first I thing I done to solve for the area I first started with the rectangle dimensions which by my drawing would be 9.5 length *5width=47.5area of rectangle. Then I took the length and width for the tip of the triangle and according to my drawing would be 4 length * 1.5 width= 6 area of triangle. Then I added the two sums together 47.5 area of rectangle + 6 area of triangle =53.5 total area of trapezoid.
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&#Good work. Let me know if you have questions. &#