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.

 

005.  `query 5

 

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?

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution: 

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

 

 

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

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

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.

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

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

 

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

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution: 

 

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

 

Thus displacement is

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.

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

 

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

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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.

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

 

 

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

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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. **

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating: 

Question: 

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

Your Solution: 

 

Confidence Rating:

 

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.

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution: 

It is about 3000 miles from coast to coast. 

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.

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

 

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?

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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.

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

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.

Your solution: 

Confidence Assessment:

Self-critique (if necessary):

Self-critique Rating:

 

 

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? 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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.. **

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

 

Add comments on any surprises or insights you experienced as a result of this assignment.

 

** I had to get a little help from a friend on vectors, but now I think I understand them. They are not as difficult to deal with as I thought.  **