If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
006. `query 6
Question: Given uniform acceleration 2 m/s^2, displacement 125 meters and final velocity 30 m/s, a student finds that the initial velocity is zero and the time interval is 10 seconds. We wish to check to see if the student's results are consistent with the given information.
Using the initial velocity and time interval obtained by the student, along with the 30 m/s final velocity, quickly reason out the acceleration and the displacement in terms of the definitions of average velocity and/or average acceleration and the assumption of uniform acceleration.
State whether the student's solutions are consistent with the originally given information.
Compare with your solution to this problem.
Your solution:
Confidence Assessment:
Given Solution:
The student obtained initial velocity zero and time interval 10 seconds. Along with the given 30 m/s final velocity this would lead to the following conclusions:
the average velocity is 15 m/s, which since acceleration is uniform is simply the average of the initial and final velocities
justification in terms of definitions: if acceleration is uniform, then the v vs t graph is linear so that the average velocity is equal to the average of initial and final velocities
multiplying the 15 m/s average velocity by the 10 s time interval interval we get displacement 150 m
justification in terms of definitions: ave velocity is ave rate of change of position with respect to clock time, so vAve = `ds / `dt, from which it follows that `ds = vAve * `dt
the change in velocity would be 30 m/s, since the velocity changes on this interval from 0 m/s to 30 m/s; so the acceleration would be aAve = `dv / `dt = 30 m/s / (10 s) = 3 m/s^2
justification in terms of definitions: ave acceleration is ave rate of change of velocity with respect to clock time, so aAve = `dv / `dt.
in this situation acceleration is uniform, so we can if we wish use just plain a instead of aAve
The student's solution is not consistent with the given information, which specified acceleration 2 m/s^2 and displacement 125 meters.
A solution to the problem:
Using the fourth equation of motion with the given information (`ds, a and vf) we have
vf^2 = v0^2 + 2 a `ds , which we solve for v0 to get
v0 = +- sqrt( vf^2 - 2 a `ds) =
+- sqrt( (30 m/s)^2 - 2 * (2 m/s^2) * (125 m) ) =
+- sqrt( 900 m^2 / s^2 - 500 m^2 / s^2) =
+- sqrt( 400 m^2 / s^2) =
+- 20 m/s.
If vf = 20 m/s then we could directly reason out the rest (vAve would be 25 m/s, so it would take 5 s to go 125 m), or we could use the first or second equation of motion to find `dt
The first equation `ds = (vf + v0) / 2 * `dt gives us `dt = 2 * `ds / (vf + v0) = 2 * 125 m / (20 m/s + 30 m/s) = 5 s.
Alternatively the second equation vf = v0 + a `dt gives us `dt = (vf - v0) / a = (30 m/s - 20 m/s) / (2 m/s^2) = 10 m/s / (2 m/s^2) = 5 s.
If vf = -20 m/s then we could directly reason out the rest (vAve would be (-20 m/s + 30 m/s) / 2 = 5 m/s, so it would take 25 s to go 125 m), or we could use the first or second equation of motion to find `dt
The first equation `ds = (vf + v0) / 2 * `dt gives us `dt = 2 * `ds / (vf + v0) = 2 * 125 m / (-20 m/s + 30 m/s) = 25 s.
Alternatively the second equation vf = v0 + a `dt gives us `dt = (vf - v0) / a = (30 m/s - (-20 m/s) ) / (2 m/s^2) = 50 m/s / (2 m/s^2) = 25 s.
STUDENT COMMENT
Umm, evidently I did NOT
understand the problem.. even looking back, I’m still not sure how
everything in the question exactly relates to the answer. I understand
the given answer and what it means, but the original
questions are very confusing to me!
INSTRUCTOR RESPONSE
The point is that the student's
solutions are inconsistent.
Using the initial velocity obtained by the student, the change in
velocity would be 30 m/s and the acceleration would be 3 m/s^2.
This of course contradicts the given acceleration, which was 2 m/s^2. So
the student's solution contradicts the given information.
STUDENT COMMENT
I do not understand this
problem at all. The information given all makes sense and I know that I
am looking for some way to
make it all fit together to verify the students results. I understand it
a little better and understand the path I was
supposed to take after reading through the explanation, but am going to
have to work on problems like these.
INSTRUCTOR RESPONSE
It would be really beneficial
for you to go through the given solution one phrase at a time, and tell
me exactly what you
do and do not understand.
For example in the first few lines:
The student obtained initial velocity zero and time interval 10 seconds. Along with the given 30 m/s final velocity this
would lead to the following conclusions:
• the average velocity is 15 m/s, which since acceleration is uniform is simply the average of the initial and final
velocities
**** do you understand how the given information leads to the conclusion that the average velocity is 15 m/s? ****
**** what do you and do you not understand about the meaning of the statement 'since acceleration is uniform is simply the
average of the initial and final velocities'? ****
the change in velocity would be 30 m/s, since the velocity changes on this interval from 0 m/s to 30 m/s
**** what do you and do you not understand about the details of this statement and its overall meaning? ****
so the acceleration would be aAve = `dv / `dt = 30 m/s / (10 s) = 3 m/s^2
**** what do you and do you not understand about the meaning of the statement acceleration is `dv / `dt? ****
**** what do you and do you not understand about why in this case `dv is 30 m/s and `dt is 10 s? ****
**** what do you and do you not understand about the calculation 30 m/s / (10 s) = 3 m/s^2? ****
You should deconstruct the entire solution, one phrase at a time, and tell me what you do and do not understand about each.
With that information I can help you address the things you don't understand.
Self-critique (if necessary):
Self-critique Rating:
Question: (gen and univ phy) At 1200 liters/day per family how much would the level of a lake with area 50 km^2 fall in a year if supplying town of population 40000? You may make your own (reasonable) assumption of how many people are in a family.
Your solution:
Confidence Assessment:
Given Solution:
You should know how the milliliter, liter and cubic meter, three common measure of volume, are related:
a milliliter is the volume of a cube 1 cm on a side
a liter is the volume of a cube 10 cm on a side
a cubic meter is the volume of a cube 1 meter on a side
so that
1 liter = (10 cm)^3 = 1000 cm^3 or 1000 milliliters
1 cubic meter = (100 cm)^2 = 1 000 000 milliliters
1 cubic meter = 1 000 000 milliliters / (1000 milliliters / liter) = 1000 liters
You should also understand the following images, which will allow you to visualize and thereby reason out these and similar relationships
It takes 10 sides of length 10 cm to make a 1 meter side, so to fill a cubic meter with 10 cm cubes we would need 10 layers, each with 10 rows and each row with 10 cubes, for a total of 10 * 10 * 10 cubes. A cubic meter is therefore 10 * 10 * 10 liters, or 1000 liters.
It takes 10 sides of length 1 cm to make a 10 cm side, so to fill a liter (a 10 cm cube) with 1 cm cubes we would need 10 layers, each with 10 rows and each row with 10 cubes, for a total of 10 * 10 * 10 cubes. A liter is therefore 10 * 10 * 10 milliliters, or 1000 milliliters.
It is also helpful to visualize the relationship between a cubic meter and a cubic kilometer:
A kilometer is 1000 meters.
A cubic kilometer is therefore (1000 meters) ^ 3 = 1 000 000 000 m^3, or 10^9 m^3, or a billion m^3.
A cubic kilometer is visualized as 1000 layers each with 1000 rows each made up of 1000 one-meter cubes, for a total of 1000 * 1000 * 1000 = 1 000 000 000 one-meter cubes.
Finally you should understand what a cylinder is and how we find its area.
The volume of a cylinder is equal to the area of its cross-section multiplied its altitude, as you saw in the q_a_initial_problems.
The links below explain prisms and cylinders, and their volumes, in elementary terms:
The solution:
A liter is 1/1000 of a cubic meter. It can be thought of as a cube 10 cm on a side. To fill a 1-meter cube t would take 10 rows with 10 cubes in each row to make a single layer 10 cm high, and 10 layers to fill the cube--i.e., 10 * 10 * 10 = 1000 one-liter cubes fill a 1-meter cube.
A km is 1000 meters so a km^2 is (1000 m)^2 = 10^6 m^2.
1200 liters / day per family, with 40,000 people at 4 persons / family, implies 10,000 families each using 1200 * .001 m^3 = 1.2 m^3 per day.
Total usage would be 10,000 families * 1.2 m^3 / day / family * 365 days / year = 4.3 * 10^6 m^3 / year.
Visualize the surface of the lake as the base of a large large irregular cylinder. The volume of a cylinder is the product of the area of its base and its altitude.
The volume of water corresponding to a depth change `dy is therefore `dy * A, where A is the area of the lake.
It might be helpful to think of a layer of ice several centimeters thick on top of the lake. The cross-sections of this layer all have very nearly the same size and shape, so it can be viewed as a cylinder with cross-sectional area equal to the area of the lake, and a thickness of several centimeters.
The area of the lake is 50 km^2 * 10^6 m^2 / km^2 = 5 * 10^7 m^2.
`dy * A = Volume so
`dy = Volume / A
= 4.3 * 10^6 m^3 / (5 * 10^7 m^2) = .086 m or 8.6 cm.
This estimate is based on 4 people per family. A different assumption would change this estimate.
STUDENT QUESTION: 40000 people in town divided by average household of 4 = 10,000 families 10,000 families * 1200 liters /day = 12000000 liters used per day * 365 days in a year = 4380000000 liter used. Here is where I get confused changing from liters to level of 50 km^2 to subtract.
INSTRUCTOR RESPONSE: If you multiply the area of the lake by the change in depth you get the volume of water used. You know the area of the lake and the volume of the water used, from which you can find the change in depth.
Of course you need to do the appropriate conversions of units. Remember that a liter is the volume of a cube 10 cm on a side, so it would take 10 rows of 10 such cubes to make one layer, then 10 layers, to fill a cube 1 meter = 100 cm on a side.
You should also see that a km^2 could be a square 1 km on a side, which would be 1000 meters on a side, to cover which would require 1000 rows of 1000 1-meters squares.
If you end of having trouble with the units or anything else please ask some specific questions and I will try to help you clarify the situation.
ANOTHER INSTRUCTOR COMMENT:
The water used can be thought of as having been spread out in a thin layer on top of that lake. That thin layer forms a cylinder whose cross-section is the surface of the lake and whose altitude is the change in the water level. The volume of that cylinder is equal to the volume of the water used by the family in a year. See if you can solve the problem from this model.
COMMON ERROR: Area is 50 km^2 * 1000 m^2 / km^2
INSTRUCTOR COMMENT:
Two things to remember: You can't cover a 1000 m x 1000 m square with 1000 1-meter squares, which would only be enough to make 1 row of 1000 squares, not 1000 rows of 1000 squares.
1 km^2 = 1000 m * 1000 m = 1,000,000 m^2. **
STUDENT QUESTION
Can you explain this part of the
problem:
The area of the lake is 50 km^2 * 10^6 m^2 / km^2 = 5 * 10^7 m^2.
Where did 10^6 m^2 / km^2 come from? I think I understand the rest of
the problem.
INSTRUCTOR RESPONSE
1 km = 1000 m, or 10^3 m.
So (1 km)^2 = (10^3 m)^2, or
1 km^2 = 10^6 m^2.
Thus 10^6 m^2 / (1 km^2) is a division of a quantity by an equal
quantity.
It follows that 10^6 m^2 / km^2 = 1
So the product 50 km^2 * 10^6 m^2 / km^2 is just 50 km^2 * 1, or 50
km^2.
Of course 50 km^2 * 10^6 m^2 / km^2 is also equal to 5 * 10^7 m^2.
Thus 50 km^2 = 5 * 10^7 m^2.</h3>
SUPPLEMENTARY QUESTIONS
These optional questions are designed to help students who still don't understand the second part of the given solution. You are welcome to copy and respond to these questions:
Assume the lake has uniform depth. If you knew that depth, how would you find the volume of water in the lake?
The area of the lake is 50 km^2. If it was 5 meters deep, how much water would there be in the lake?
How deep would the lake have to be to contain 2 * 10^7 m^3 of water?
How deep would the lake have to be to contain 4 * 10^6 m^3 of water?
Here's the second part of the solution:
Visualize the surface of the lake as the base of a large large irregular
cylinder. The volume of a cylinder is the product of the area of its
base and its altitude.
• The volume of water corresponding to a depth change `dy is therefore `dy
* A, where A is the area of the lake.
• It might be helpful to think of a layer of ice several centimeters
thick on top of the lake. The cross-sections of this layer all have very
nearly the same size and shape, so it can be viewed as a cylinder with
cross-sectional area equal to the area of the lake, and a thickness of
several centimeters.
The area of the lake is 50 km^2 * 10^6 m^2 / km^2 = 5 * 10^7 m^2.
`dy * A = Volume so
• `dy = Volume / A
= 4.3 * 10^6 m^3 / (5 * 10^7 m^2) = .086 m or 8.6 cm.
Having worked on the questions I've inserted, you should have a better
understanding of the given solution.
What do you now understand and what do you not understand about the
given solution?
Self-critique (if necessary):
Self-critique Rating:
Question:
An automobile coasts, starting at high speed, up a hill, slowing as it goes. It encounters significant air resistance, which causes it to slow more quickly than it would in the absence of air resistance.
If the positive direction is down the hill, then
If the positive direction is up the hill, then
Your solution:
Confidence Assessment:
Self-critique (if necessary):
Self-critique Rating:
Question: univ 1.74 (11th edition 1.70) univ sailor 2 km east, 3.5 km SE, then unknown, ends up 5.8 km east find magnitude and direction of 3d leg, explain how diagram shows qualitative agreement
Your solution:
Confidence Assessment:
Given Solution:
`a** Letting the three vectors be A, B and C (C unknown) and the x axis point east we have A at 0 degrees, B at 315 degrees and the resultant at 0 degrees.
Ax = 2, Ay = 0 (A is toward the East, along the x axis).
Bx = B cos(315 deg) = 2.47, By = B sin(315 deg) = -2.47.
Rx = 5.8, Ry = 0.
Since Ax + Bx + Cx = Rx we have Cx = Rx - Ax - Bx = 5.8 - 2 - 2.47 = 1.33.
Since Ay + By + Cy = Ry we have Cy = Ry - Ay - By = 0 - 0 - (-2.47) = 2.47.
C has magnitude sqrt( Cx^2 + Cy^2) = sqrt ( 1.33^2 + 2.47^2) = sqrt(7.9) = 2.8, representing 2.8 km.
C has angle arctan(Cy / Cx) = arctan(2.47 / 1.33) = 61 deg. **
STUDENT QUESTION
Why is it 315 deg and not 45? Would you write it 315 deg northwest of east since it goes in that direction?
INSTRUCTOR RESPONSE
With the x and y axes in
standard position, with the x axis pointing east and the y axis north,
the southeasterly direction lies in the fourth quadrant, at 315 degrees
as measured counterclockwise from the positive x axis.
If you measure your vectors from anywhere else you can't use the simple
relationships Ax = A cos(theta) and Ay = A sin(theta). With this
convention you can, and the signs of the trig functions automatically
take care of the + and - signs of the components.
You should of course also be able to use triangle trigonometry, but the
circular trigonometry of this solution will be used in most of the
solutions you will see in the queries and qa's.
STUDENT COMMENT/QUESTION:
I had this totally wrong. I am
sooooo confused as the student above was why the B vector would be 315
degrees? How come it is in the 4th quadrant? I put my vector going in
the SE direction. Why is this incorrect?
INSTRUCTOR RESPONSE
You could orient your x-y
coordinate system in any way you wish, as long as the positive y axis is
at 90 degrees counterclockwise relative to the positive x axis.
You could orient the system so that the easterly direction is the y
direction, which would be consistent with your 90 degree angle for the 2
km vector. If so, the x direction would have to be toward the
south. This would put the southeasterly direction in the first quadrant.
The southeasterly direction would be at 45 degrees with this
orientation.
The given solution assumes the x axis to be pointing east, so the y axis
points north, and the southeasterly direction is 'below' the x axis, in
the 4th quadrant, at 315 degrees.
Self-critique (if necessary):
Self-critique Rating:
Question: `q**** query univ 1.86 (11th edition 1.82) (1.66 10th edition) vectors 3.6 at 70 deg, 2.4 at 210 deg find scalar and vector product
Your solution:
Confidence Assessment:
Given Solution:
`a** GOOD STUDENT SOLUTION WITH INSTRUCTOR COMMENT:
A * B = |A||B| cos(theta) = AxBx + Ay By + AzBz
3.6 * 2.4 * cos (140 deg) = -6.62
To check for consistency we can calculate the components of A and B:
Ax = 3.6 * cos(70 deg) = 1.23
Ay = 3.6 * sin(70 deg) = 3.38
Bx = 2.4 * cos (210 deg) = -2.08
By = 2.4 * sin(210 deg) = -1.2
dot product = 1.23 *-2.08 + 3.38 * -1.2 = -2.5584 + -4.056 = -6.61. Close enough.
----------------------
Cross product:
| A X B | = |A| |B| sin (theta) = 3.6 * 2.4 * sin(140) = 5.554.
Finding the components we have
(Ay * Bz - Az * By) i + (Az * Bx - Ax + Bz) j + (Ax * By - Ay * Bx) k =
((3.38*0)-(0*-1.2)] i + [(0 * -2.08) - (1.23 * 0)] j + [(1.23*-1.2)-(3.38 * -2.08)] k =
0 i + 0 j + 5.55 k,
or just 5.55 k, along the positive z axis ('upward' from the plane).
INSTRUCTOR COMMENT: Your conclusion is correct. The cross product must be at a right angle to the two vectors. To tell whether the vector is 'up' or 'down' you can use your result, 5.55 k, to see that it's a positive multiple of k and therefore upward.
The other way is to use the right-hand rule. You place the fingers of your right hand along the first vector and position your hand so that it would 'turn' the first vector in the direction of the second. That will have your thumb pointing upward, in agreement with your calculated result, which showed the cross product as being in the upward direction. **
Self-critique (if necessary):
Self-critique Rating: