Query 05

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course Phy 231

2/4 11:50pm

005. `query 5

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Question: `qIntro Prob 6 given init vel, accel, `dt find final vel, displ

If initial velocity is v0, acceleration is a and time interval is `dt, then in terms of these three symbols what are the final velocity vf and the displacement `ds?

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Your solution:

We first use a and `dt to find `dv:

a=`dv/`dt

(a)(`dt)=`dv

Therefore: vf = v0+(a)(`dt)

Then we use two definitions of vAve:

vAve=(v0+vf)/2

and vAve=(`ds)/(`dt)

Combine them to find:

(`ds)/(`dt)=(v0+vf)/2

`ds=(`dt)[(v0+vf)/2]

`ds=(`dt)[.5(v0+vf)]

Now we need to use our understanding of acceleration to eliminate the vf in that equation.

If a=(vf-v0)/(`dt),

then a(`dt)=vf-v0, and vf= a(`dt)+v0

Plug that into the equation above:

`ds=(`dt)[.5(v0+a(`dt)+v0]

`ds=(`dt)[.5(2v0 + a(`dt)]

`ds=(`dt)(v0 + .5a(dt))

Or: `ds =(v0*`dt) + .5a(`dt^2)

confidence rating #$&*: 3

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Given Solution:

`a**You would use accel. and `dt to find `dv:

a * `dt = `dv.

Adding `dv to initial vel. v0 you get final vel.

Then average initial vel. and final vel. to get ave. vel.:

(v0 + vf) / 2 = ave. vel.

You would then multiply ave. vel. and `dt together to get the displacement

For example if a = 3 m/s^2, `dt = 5 s and v0 = 3 m/s:

3 m/s^2 * 5 s = 15 m/s = `dv

15 m/s + 3 m/s = 18 m/s = fin. vel.

(18 m/s + 3 m/s) / 2 = 10.5 m/s = vAve

10.5 m/s * 5 s = 52.5 m = displacement

In more abbreviated form:

a * `dt = `dv

v0 + `dv = vf

(vf + v0) /2 = vAve

vAve * `dt = `ds so

`ds = (vf + v0) / 2 * `dt.

**

STUDENT QUESTION

If we have the formula vf= v0 + a * dt, then we would substract the v0 from both sides to isolate the a * dt algebraically, so our formula would be vf-v0= a* `dt,

how is this in comparison to the initial velocity v0 + the change in velocity(dv) = to the final velocity(vf).

If we multiply the acceleration(a) times time(dt) we find the change in velocity(dv).......we then add the initial to the change to find the final.......

Why do we add the initial to the change in velocity to find the final?

INSTRUCTOR RESPONSE

The initial velocity is v0, the final velocity is vf, so the change in velocity is `dv = vf - v0.

Thus your early result vf-v0= a* `dt shows that a * `dt is equal to `dv.

In general the change in any quantity is equal to its final value minus its initial value.

It follows immediately from this that if you add the change in the quantity to its original value, you get its final value.

The following two statements say the same thing:

statement 1: If the temperature starts at 20 degrees and ends up at 35 degrees then it changed by +15 degrees.

statement 2: If the temperature starts at 20 degrees and changes by +15 degrees then it ends up at 35 degrees.

We generalize this to the two symbolic statements

If a quantity Q changes from Q0 to Qf then the change is `dQ = Qf - Q0.

If a quantity Q starts out at Q0 and changes by `dQ, then it ends up at Qf.

These statements can be expressed as two equations

`dQ = Qf - Q0 and

Qf = Q0 + `dQ

These two equations are algebraically equivalent: you can get the second by adding Q0 to both sides of the first, or you can get the first by subtracting Q0 from both sides and reversing sides.

A third equation also follows:

Q0 = Qf - `dQ,

which can be interpreted in terms of the preceding examples into obvious statements.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in time interval `dt?

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Your solution:

vAve=`ds/`dt

so, (v0+vf)/2=`ds/`dt

and (v0+vf)/2 * `dt =`ds

confidence rating #$&*:

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Given Solution:

Since accel is uniform vAve = (v0 + vf) / 2.

Thus displacement is

`ds = vAve * `dt = (v0 + vf) / 2 * `dt,

which is the first equation of uniformly accelerated motion. **

STUDENT QUESTION

I failed to make reference to uniformly accelerated motion.

What exactly is the difference between uniformly accelerated motion and average acceleration??? Will we be asked to

differentiate between the two for problems, or is this something we should be able to determine on our own easily???

INSTRUCTOR RESPONSE

Uniformly accelerated motion is motion in which the acceleration is uniform, unchanging.

If motion is uniformly accelerated, then the acceleration is constant, so the acceleration at any instant is equal to the average acceleration.

If motion is uniformly accelerated, then since the slope of the velocity vs. clock time graph represents acceleration, the slope is constant; i.e., the graph is a straight line.

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Self-critique (if necessary):

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Question: Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.

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Your solution:

We begin with v0, vf, and `dt on the first level, since they are given in the premise of the problem.

The second level uses v0 and vf to easily find v_Ave and `dv.

Then we can draw from `dt on the first line and `dv on the second line to determine aAve on the third line. We can also draw from `dt on the first line and vAve on the second line to determine `ds on the third line.

confidence rating #$&*:

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Given Solution:

`a** The first level in the diagram would contain `dt, v0 and vf.

From v0 and vf we can easily reason out `dv, so v0 and vf would connect to `dv in the second level.

The second level would also contain vAve, also obtained from v0 and vf and therefore connected from vf in the first level to v0 in the first level.

The third level would contain an a, which is reasoned out from `dv and `dt and so is connected to `dv in the second level and `dt in the first level.

The third level would also contain `ds, which follows from vAve and `dt and is therefore connected to vAve in the second level and `dt in the first level. **

STUDENT QUESTION:

I'm not sure what is meant by a flow diagram. I know that we can determine 'ds from the equation 'ds=(v0+vf)/2* 'dt. Then I can use 'ds to find other possible information by plugging this and other information into other equations.

INSTRUCTOR RESPONSE

The instructor's response developed into an entire document, a bit too long to include in this query without interrupting the flow. The document has been posted at

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/flow_diagrams.htm

and should be very useful to anyone who is having trouble with the idea of flow diagrams.

STUDENT COMMENT

Flow diagrams are useful in that they give us something to logically grind out. It's not enough to know that there are formulas to find variables. True learning is when a person can take whats given, twist it and manipulate it, and find other answers.

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Self-critique (if necessary):

This concept seemed more symbolic than I typically tend to think things through, but after doing it a few times I appreciate the idea of an organized way to start by stating what I already know and deducing things step by step from there.

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Self-critique Rating:

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Question: `qDescribe the flow diagram we obtain for the situation in which we know v0, vf and `ds.

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Your solution:

We write v0, vf and `ds on the first level.

We draw two sets of lines from v0 and vf down to the second level, where we connect to vAve and `dv, both easily determined from initial and final velocity.

Drawing from `ds on the first level and vAve on the second level, we can determine `dt and place it on the third level.

We can then draw from `dv on the second level and `dt on the third level to determine aAve, and place it on the third level as well.

confidence rating #$&*: 3

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Given Solution:

`a** The flow diagram shows us the flow of information, what we get from what, usually by combining two quantities at a time. How we get each quantity may also be included.

From vf and v0 we get `dv, shown by lines from vf and v0 at the top level to `dv. From vf and v0 we also get and vAve, shown by similar lines running from v0 and vf to vAve.

Then from vAve and `ds we get `dt, with the accompanying lines indicating from vAve and `ds to `dt.

Then from `dv and our newly found `dt we get acceleration, indicated similarly. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question:

Suppose we have two points on a straight-line graph of velocity vs. clock time.

How do we construct a trapezoid to represent the motion on the intervening interval?

What aspect of the graph represents the change in velocity for the interval, and why?

What aspect of the graph represents the change in clock time for the interval, and why?

What aspect of the graph represents the acceleration for the interval, and why?

What aspect of the graph represents the displacement for the given interval, and why?

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Your Solution:

We construct a trapezoid by drawing vertical lines from each of the two points at the ends of the straight-line graph down to the x-axis.

The difference in the height of the two points represents the change in velocity. Because velocity is measured on the y-axis, the y-coordinate for each of those points tells us the velocity at that time. As velocity changes over the time interval, the value of y goes up or down. The difference between the two y values represents the difference in velocity.

The span from the two points on the x-axis directly below the points on our straight-line graph represents the change in clock time. Since we are measuring time on the x-axis, the x coordinate of each of those two points marks the times at which we are given initial and final velocity. The space between them (or the difference between them) is the time span.

The slope of the straight-line graph represents the acceleration. This is because the slope tells us about the relationship between the rise and the run (in this case, the rate of change in velocity with respect to time). That relationship is the acceleration.

The area formed below the sloped line, with the two vertical lines as side boundaries, tells us the displacement. If we know how fast something is traveling, on average, and we know how long it is doing so, we can figure out how far it goes by multiplying the two known quantities. Here, that is represented by finding the area of the trapezoid using average height (average velocity) and multiplying it by the width of the base (elapsed time).

confidence rating #$&*:

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Question: `qPrinciples of Physics and General College Physics Students: Prob. 1.26: Estimate how long it would take a runner at 10 km / hr to run from New York to California. Explain your solution thoroughly.

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Your solution:

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Given Solution:

It is about 3000 miles from coast to coast.

A km is about .62 mile, so 3000 miles * 1 km / (.62 miles) = 5000 km, approximately.

At 10 km / hr, the time required would be

5000 km / (10 km / hr) =

500 km / (km/hr) =

500 km * (hr / km) =

500 (km / km) * hr =

500 hr.

Be sure you understand the units of this calculation. Units should be used at every step of every calculation.

The corresponding symbolic solution:

vAve = `ds / `dt; we want to find `dt so we solve to get

`dt = `ds / vAve. Substituting `ds = 5000 km and vAve = 10 km/hr we have

`dt = 5000 km / (10 km/hr) = 500 hr.

STUDENT SOLUTION (with some inconsistencies in units)

The student's estimate of the distance was 4000 km, which is perfectly OK:

To find out how much time it takes to travel this far, I took 4000 km and divided it by 10 km/h.

This was set up as follows:

4000 km / 10 km

This becomes 400 km * 1 hr

Our kilometers cancel out and we are left with 400 hours to run from New York to California.

INSTRUCTOR RESPONSE

I would certainly accept your solution, with little or no penalty at the level of Phy 121.

However your use of units does have some contradictions, and you will understand units better if you understand them:

In the first place, 4000 km / (10 km) = 400, not 400 km. The km divide out.

400 represented the number of 10 km intervals in a 4000 km trip.

Since average speed is 10 km/hr, meaning that a 10 km interval is covered each hour, it therefore takes about 400 hours to complete the trip.

Note also that the calculation given in your solution as 400 km * 1 hr would be 400 km * hr, not the 400 hr you intend.

Finally, to use the fact that v_Ave = `ds / `dt:

The time to cover distance `ds at average speed v_Ave is `dt = `ds / v_Ave, and that the units of v_Ave are km / hr. So to be entirely correct, the correct calculation could read

`dt = `ds / v_Ave = 4000 km / (10 km/hr) = 400 hr.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: All Students: Estimate the number heartbeats in a lifetime. What assumptions did you make to estimate the number of heartbeats in a human lifetime, and how did you obtain your final result?

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Your solution: If an average heartbeat is about 70 beats per minute, then it beats

70*60=4200 times in an hour,

4200*24=100800 times in a day,

100800*365=36,792,000 times in a year

36,792,000*75= 2,759,400,000 times in a 75-year lifetime.

So, about 2.8 billion beats.

confidence rating #$&*: 3

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Given Solution:

Typical assumptions:

At 70 heartbeats per minute, with a lifetime of 80 years, we have

70 beats / minute * 60 minutes/hour * 24 hours / day * 365 days / year * 80 years =

3 billion beats, approximately.

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Self-critique (if necessary):

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Question: Describe the reasoning process you would use in a situation where you are given v0, a and `dt.

Describe the flow diagram that represents your reasoning.

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Your solution:

On the top level, we write v0, a, and `dt.

Drawing from a and `dt, we can determine `dv and place it on the second level.

Then we can draw from v0 and `dv to easily determine vf and place it on the second level as well.

Next, we can connect v0 and vf and draw down to the third level, where we can determine vAve.

Finally, we can draw down from `dt and connect it with vAve, placing `ds on the third level as well.

confidence rating #$&*: 3

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Self-critique (if necessary):

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Question: University Physics Students Only: Problem 1.55 (11th edition 1.52) (i.e., Chapter 1, Problem 52): Angle between -2i+6j and 2i - 3j. What angle did you obtain between the two vectors?

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Your solution:

Angle between A and y-axis = arctan(2/6) = 18.43deg

Angle between B and x-axis = arctan(3/2) = 56.31deg

Angle between x-axis and y-axis = 90 deg

total = 164.74 deg

confidence rating #$&*: 2

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Given Solution:

For the given vectors we have

dot product =-2 * 2 + 6 * (-3) = -22

magnitude of first vector = sqrt( (-2)^2 + 6^2) = sqrt(40)

magnitude of second vector = sqrt( 2^2 + (-3)^2 ) = sqrt(13)

Since dot product = magnitude of 1 st vector * magnitude of 2d vector * cos(theta) we have

cos(theta) = dot product / (magnitude of 1 st vector * magnitude of 2d vector)

so that

theta = arccos [ dot product / (magnitude of 1 st vector * magnitude of 2d vector) ]

= arccos[ -22 / ( sqrt(40) * sqrt(13) ) ] = arccos ( -.965) = 164 degrees, approx.. **

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Self-critique (if necessary):

I need some more practice on dot products, because my instinct was to approach this problem totally differently; I sketched out the two vectors based on knowing their components. Since vector A formed an angle to the left of the y-axis that I could determine and vector B formed an angle below the x-axis that I could determine, I added those two angles plus 90 degrees between the axes.

Reading through your explanation makes sense to me but it's not coming naturally to me yet. It didn't occur to me that finding the magnitude of either vector needed to be involved in the solution when we were only concerned with angles. But I know I need to be familiar with this process for the future.

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Self-critique Rating: 2

@&

Your method would not work with vectors in 3 dimensions, so it's essential to be able to do this in terms of the dot product.

*@

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Question: University Physics Students Only: Problem 1.55 (11th edition 1.52) (i.e., Chapter 1, Problem 52): Angle between -2i+6j and 2i - 3j. What angle did you obtain between the two vectors?

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Your solution:

Angle between A and y-axis = arctan(2/6) = 18.43deg

Angle between B and x-axis = arctan(3/2) = 56.31deg

Angle between x-axis and y-axis = 90 deg

total = 164.74 deg

confidence rating #$&*: 2

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Given Solution:

For the given vectors we have

dot product =-2 * 2 + 6 * (-3) = -22

magnitude of first vector = sqrt( (-2)^2 + 6^2) = sqrt(40)

magnitude of second vector = sqrt( 2^2 + (-3)^2 ) = sqrt(13)

Since dot product = magnitude of 1 st vector * magnitude of 2d vector * cos(theta) we have

cos(theta) = dot product / (magnitude of 1 st vector * magnitude of 2d vector)

so that

theta = arccos [ dot product / (magnitude of 1 st vector * magnitude of 2d vector) ]

= arccos[ -22 / ( sqrt(40) * sqrt(13) ) ] = arccos ( -.965) = 164 degrees, approx.. **

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Self-critique (if necessary):

I need some more practice on dot products, because my instinct was to approach this problem totally differently; I sketched out the two vectors based on knowing their components. Since vector A formed an angle to the left of the y-axis that I could determine and vector B formed an angle below the x-axis that I could determine, I added those two angles plus 90 degrees between the axes.

Reading through your explanation makes sense to me but it's not coming naturally to me yet. It didn't occur to me that finding the magnitude of either vector needed to be involved in the solution when we were only concerned with angles. But I know I need to be familiar with this process for the future.

------------------------------------------------

Self-critique Rating: 2

@&

Your method would not work with vectors in 3 dimensions, so it's essential to be able to do this in terms of the dot product.

*@

#*&!

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Question: University Physics Students Only: Problem 1.55 (11th edition 1.52) (i.e., Chapter 1, Problem 52): Angle between -2i+6j and 2i - 3j. What angle did you obtain between the two vectors?

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Your solution:

Angle between A and y-axis = arctan(2/6) = 18.43deg

Angle between B and x-axis = arctan(3/2) = 56.31deg

Angle between x-axis and y-axis = 90 deg

total = 164.74 deg

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

For the given vectors we have

dot product =-2 * 2 + 6 * (-3) = -22

magnitude of first vector = sqrt( (-2)^2 + 6^2) = sqrt(40)

magnitude of second vector = sqrt( 2^2 + (-3)^2 ) = sqrt(13)

Since dot product = magnitude of 1 st vector * magnitude of 2d vector * cos(theta) we have

cos(theta) = dot product / (magnitude of 1 st vector * magnitude of 2d vector)

so that

theta = arccos [ dot product / (magnitude of 1 st vector * magnitude of 2d vector) ]

= arccos[ -22 / ( sqrt(40) * sqrt(13) ) ] = arccos ( -.965) = 164 degrees, approx.. **

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Self-critique (if necessary):

I need some more practice on dot products, because my instinct was to approach this problem totally differently; I sketched out the two vectors based on knowing their components. Since vector A formed an angle to the left of the y-axis that I could determine and vector B formed an angle below the x-axis that I could determine, I added those two angles plus 90 degrees between the axes.

Reading through your explanation makes sense to me but it's not coming naturally to me yet. It didn't occur to me that finding the magnitude of either vector needed to be involved in the solution when we were only concerned with angles. But I know I need to be familiar with this process for the future.

------------------------------------------------

Self-critique Rating: 2

@&

Your method would not work with vectors in 3 dimensions, so it's essential to be able to do this in terms of the dot product.

*@

#*&!#*&!

&#Good work. See my notes and let me know if you have questions. &#