Open QA 6

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course phy 121

If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://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.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.

006. Using equations with uniformly accelerated motion.

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Question: `q001. Note that there are 12 questions in this assignment.

Using the equation vf = v0 + a * `dt determine the acceleration of an object whose velocity increases at a uniform rate from 10 m/s to 30 m/s in 15 seconds. Begin by solving the equation for the acceleration a, then 'plug in' your initial and final velocities. Describe your work step y step.

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Your solution:

The acceleration is equal to (vf-v0) / `dt or the change in velocity which is (30cm/s - 10cm/s = 20m/s) / `dt (15 sec) = 1.3m/s/s

confidence rating #$&*:

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Given Solution:

The equation vf = v0 + a * `dt is solved for a by first adding -v0 to both sides to obtain vf - v0 = v0 + a * `dt - v0, which simplifies to vf - v0 = a * `dt. Both sides are then divided by `dt to obtain (vf - v0) / `dt = a. Reversing left-and right-hand sides we obtain the formula a = (vf - v0) / `dt.

We then plug in our given values of initial and final velocities and the time interval. Since velocity increases from 10 m/s to 30 m/s, initial velocity is v0 = 10 m/s and final velocity is vf = 30 m/s. The time interval `dt is 15 seconds, so we have a = (30 m/s - 10 m/s) / (15 s) = 20 m/s / (15 s) = 1.33.. m/s^2.

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Self-critique (if necessary):

I got the correct answer but I wasn’t as successfully able to break down the algebra like you did. I know what the formula is supposed to look like and how it corresponds to the most intuitive one, how important is it that I be able to do this?

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Self-critique rating:

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You do have an excellent grasp of these ideas. The formulas help you extend that.

To solve many of the problems that arise in this course you do want to be able to do the algebra of these equations.

We have

vf = v0 + a * `dt

Subtracting v0 from both sides we have

vf - v0 = v0 + a * `dt - v0

which simplifies to

vf - v0 = a * `dt.

Dividing both sides by `dt we have

(vf - v0) / `dt = a * `dt / `dt

which simplifies to

(vf - v0) / `dt = a.
*@

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Question: `q002. It wasn't necessary to use an equation to solve this problem. How could this problem had been reasoned out without the use of an equation?

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Your solution:

We could reason that if an object goes from 10m/s to 30 m/s it has changed velocity by 20 m/s. If we want to know the acceleration we divide the change in velocity by the change in time, which we know to be 15 sec to give us our answer. We could also have found that average velocity by adding the initial and final velocities and dividing by 2.

confidence rating #$&*:

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Given Solution:

Knowing that acceleration is the average rate at which velocity changes, we would first find the change in velocity from 10 meters/second to 30 meters/second, which is 20 meters/second. We would then divided change in velocity by the time interval to get 20 meters/second / (15 sec) = 1.33 m/s^2.

STUDENT QUESTION (about reasoning vs. using the equation)

I understand but the steps taken to get to the acceleration were the steps of the equation?????

INSTRUCTOR RESPONSE

The steps outlined here are the steps we could use to derive the equation. However it's possible to use the equation blindly, without understanding the reasoning behind it. In fact this is how most student use the equation, if not asked questions of this nature about the reasoning.

So, this question asks for the reasoning.

The first statement in the given solution is

'Knowing that acceleration is the average rate at which velocity changes, we would first find the change in velocity from 10 meters/second to 30 meters/second, which is 20 meters/second.'

When using the equation you never explicitly find or reason out the change in velocity, though of course the change in velocity is there in the equation, represented by the term a * `dt. In other words, you do find it, but you can use the equation without ever recognizing that you have done so.

Similarly the step a = (30 m/s - 10 m/s) / 15 s in your equation-based solution does correctly divide the change in velocity by the time interval, but you can use the equation to do this without ever recognizing that you have done so.

The direct reasoning solution never mentions or uses the equation, though of course direct reasoning can be used to derive the equation.

This should help illustrate the difference between direct reasoning and using an equation. Both skills are important.

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Self-critique (if necessary): ok

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Self-critique rating: ok

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Question: `q003. Use the equation `ds = (vf + v0) / 2 * `dt to determine the initial velocity of an object which accelerates uniformly through a distance of 80 meters in 10 seconds, ending up at a velocity of 6 meters / sec. begin by solving the equation for the desired quantity. Show every step of your solution.

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Your solution:

We can change this equation algebraically which is currently a painstaking task given my semi-faded algebra background, nevertheless, for the equation v0= 2*`ds/`dt-vf we get

= 2 * (80m) / 10sec - 6m/sec

= 160m/ ????here I cannot figure out what to do, I cant subtract 10 sec from 6m/sec because they are different units. In the solution you have them being added to get 16 m/sec. How did this happen????

confidence rating #$&*:

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Given Solution:

We begin by solving the equation for v0. Starting with

`ds = (vf + v0) / 2 * `dt, we can first multiply both sides of the equation by 2 / `dt, which gives us

`ds * 2 / `dt = (vf + v0) / 2 * `dt * 2 / `dt. The right-hand side can be rearranged to give

(vf + v0) * `dt / `dt * 2 / 2; since `dt / `dt = 1 and 2 / 2 = 1

the right-hand side becomes just vf + v0. The equation therefore becomes

2 * `ds / `dt = vf + v0. Adding -vf to both sides we obtain

v0 = 2 * `ds / `dt - vf.

We now plug in `ds = 80 meters, `dt = 10 sec and vf = 6 m/s to get

v0 = 2 * 80 meters / 10 sec - 6 m/s = 160 meters / 10 sec - 6 m/s = 16 m/s - 6 m/s = 10 m/s.

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Self-critique (if necessary): This implies that the initial velocity is greater than the final velocity which is something we haven’t really seen all that much but I can certainly see how it is appropriate. My biggest hang up at the moment is the algebra involved rearranging the equations. Im very uncertain about my methods for doing so and am relying solely on your solutions to determine the proper equation. I am hoping that in time it will become clearer but I’m getting a lot of curve balls, especially with the 10sec-6cm/s= 16 cm/s

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Self-critique rating:

@&
You're on the right track; just remember the order of operations.

2 * (80m) / 10sec - 6m/sec =
160 m / 10 sec - 6 m/s.

By order of operations we do the division before the subtraction, which gives us

16 m / s - 6 m/s.

Those expressions are compatible for subtraction. We get the final result 10 m/s.

Algebra skills, incidentally, should come back with the practice you'll be getting. The algebra for your course doesn't get that complicated; really no more complicated than what you've already seen here.
*@

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Question: `q004. We can reconcile the above solution with straightforward reasoning. How could the initial velocity have been reasoned out from the given information without the use of an equation? Hint: two of the quantities given in the problem can be combined to give another important quantity, which can then be combined with the third given quantity to reason out the final velocity.

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Your solution:

In this case we can determine the average velocity by dividing the change in position by the change in time, which gives us 8 m/sec. With a final velocity of 6m/s we can deduce that for the average velocity to be 8m/sec, the sum of the initial and final velocity divided by 2 is the vAve. (10cm/s + 6cm/s) / 2 = 8m/sec

confidence rating #$&*:

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Given Solution:

The average velocity of the object is the average rate at which its position changes, which is equal to the 80 meters change in position divided by the 10 s change in clock time, or 80 meters / 10 sec = 8 meters / sec. Since the 8 m/s average velocity is equal to the average of the unknown initial velocity and the 6 m/s final velocity, we ask what quantity when average with 6 m/s will give us 8 m/s. Knowing that the average must be halfway between the two numbers being averaged, we see that the initial velocity must be 10 m/s. That is, 8 m/s is halfway between 6 m/s and 10 m/s.

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Self-critique (if necessary): ok

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Self-critique rating: ok

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Question: `q005. Using the equation `ds = v0 `dt + .5 a `dt^2 determine the initial velocity of an object which accelerates uniformly at -2 m/s^2, starting at some unknown velocity, and is displaced 80 meters in 10 seconds. Begin by solving the equation for the unknown quantity and show every step.

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Your solution:

Here we change the equation by subtracting 0.5a*`dt^2 from both sides giving us

v0 -`dt = `ds - 0.5 (a * `dt^2). Dividing by `dt gives us

v0 = (`ds - 0.5a * `dt^2) / `dt

From here we insert the units given to us

= 80 m - 0.5 * (-2 m/s^2) * (10sec)^2 / 10 sec

= 80 m - 0.5 * (-2 m/s^2) * 100sec^2 / 10 sec

= 80 m - 0.5 * (-200m) / 10 sec

= 80 m -(-100) / 10 sec

= 180 m / 10 sec

= 18 m / sec

confidence rating #$&*:

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Given Solution:

The unknown quantity is the initial velocity v0. To solve for v0 we start with

`ds = v0 `dt + .5 a `dt^2. We first add -.5 a `dt^2 to both sides to obtain

`ds - .5 a `dt^2 = v0 `dt. We then divide both sides by `dt to obtain

(`ds - .5 a `dt^2) / `dt = v0.

Then we substitute the given displacement `ds = 80 meters, acceleration a = -2 m/s^2 and time interval `dt = 10 seconds to obtain

v0 = [ 80 meters - .5 * (-2 m/s^2) * (10 sec)^2 ] / (10 sec)

= [ 80 meters - .5 * (-2 m/s^2) * 100 s^2 ] / (10 sec)

= [ 80 m - (-100 m) ] / (10 sec)

= 180 m / (10 s) = 18 m/s.

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Self-critique (if necessary):

I had to follow the solution on that one and work it out on paper until it made sense but it is becoming clearer to me now.

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Self-critique rating:

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Question: `q006. Check the consistency of this result by verifying, by direct reasoning rather than equations, that an object whose initial velocity is 18 m/s and which accelerates for 10 seconds at an acceleration of -2 m/s^2 does indeed experience a displacement of 80 meters.

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Your solution:

If from the initial velocity of 18 m/s the object accelerates for 10 sec at -2 m/s^2 then for every 1 of the 10 seconds the 18 m/s decreases by 2m/s. The velocity will therefore change by -2m/s^2 * 10 s = -20m/s If we take the change in velocity and subtract it from the initial velocity we will obtain our final velocity of -2m/s. From this we can obtain our average velocity (18m/s +(-2m/s) / 2 = 8m/s. The product of the average velocity and the change in time gives us our distance of 80 meters. The solution is correct.

confidence rating #$&*:

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Given Solution:

The change in the velocity of the object will be -2 m/s^2 * 10 s = -20 m/s.

The object will therefore have a final velocity of 18 m/s - 20 m/s = -2 m/s.

Its average velocity will be the average (18 m/s + (-2 m/s) ) / 2 = 8 m/s.

An object which travels at an average velocity of 8 m/s for 10 sec will travel 80 meters.

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Self-critique (if necessary): ok

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Self-critique rating: ok

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Question: `q007. Using the equation vf^2 = v0^2 + 2 a `ds determine the initial velocity of an object which attains a final velocity of 20 meters/second after accelerating uniformly at 2 meters/second^2 through a displacement of 80 meters. Begin by solving the equation for the unknown quantity and show every step.

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Your solution:

Using our 4th fundamental equation we obtain the following

Vf^2=v0^2+2(a*`ds) subtract -2 a * `ds

Vf^2 -2 a * `ds = v0^2 to remove the exponent from the initial velocity we take the square root of both sides

V0= +- `sqrt (vf^2 - 2 a `ds) now we are ready to solve for the quantities we have given and discover the initial velocity

= +- `sqrt [(20m/s)^2 - 2 (2m/s^2 * 80m)]

= +- `sqrt [(400m^2/s^2) - 2 (2m/s^2 * 80m)]

= +- `sqrt [(400m^2/s^2) - 2 (160m^2/s^2)]

= +- `sqrt [(400m^2/s^2) - 320m^2/s^2)]

= +- `sqrt (80m^2/s^2)

= 8.9 m/s

confidence rating #$&*:

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Given Solution:

To solve for the unknown initial velocity v0 we start with

vf^2 = v0^2 + 2 a `ds. We first add -2 a `ds to both sides to obtain

vf^2 - 2 a `ds = v0^2. We then reverse the right-and left-hand sides and take the square root of both sides, obtaining

v0 = +- `sqrt( vf^2 - 2 a `ds).

We then substitute the given quantities vf = 20 m/s, `ds = 80 m and a = 3 m/s^2 to obtain

v0 = +- `sqrt( (20 m/s)^2 - 2 * 2 m/s^2 * 80 m) = +- `sqrt( 400 m^2 / s^2 - 320 m^2 / s^2) = +- `sqrt(80 m^2 / s^2) = +- 8.9 m/s (approx.).

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Self-critique (if necessary): ok

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Self-critique rating: ok

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Question: `q008. We can verify that starting at +8.9 m/s an object which attains a final velocity of 20 m/s while displacing 80 meters must accelerate at 2 m/s^2. In this case the average velocity will be ( 8.9 m/s + 20 m/s) / 2 = 14.5 m/s (approx) and the change in velocity will be 20 m/s - 8.9 m/s = 11.1 m/s.

At average velocity 14.5 meters/second the time required to displace the 80 meters will be 80 m / (14.5 sec) = 5.5 sec (approx).

The velocity change of 11.1 meters/second in 5.5 sec implies an average acceleration of 11.1 m/s / (5.5 sec) = 2 m/s^2 (approx), consistent with our results.

Verify that starting at -8.9 m/s the object will also have acceleration 2 meters/second^2 if it ends up at velocity 20 m/s while displacing 80 meters.

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Your solution:

If the initial velocity is -8.9 m/s then the average velocity is

vAve = (-8.9m/s + 20m/s) / 2 = 5.55 m/s The change in velocity will be figured by

`dv = (20m/s - (- 8.9m/s) = 28.9 m/s

With the vAve we can determine the change in time `ds (80 m) / vAve (5.55m/s) = 14.41sec

Now we can say that the acceleration is aAve= `dv ( 28.9m/s) / `dt (14.41 sec) = 2 m/s^2

confidence rating #$&*:

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Given Solution:

In this case the average velocity will be ( -8.9 m/s + 20 m/s) / 2 = 5.5 m/s (approx) and the change in velocity will be 20 m/s - (-8.9 m/s) = 28.9 m/s (approx). At average velocity 5.5 meters/second the time required to displace the 80 meters will be 80 m / (5.5 sec) = 14.5 sec (approx). The velocity change of 28.5 meters/second in 14.5 sec implies an average acceleration of 28.5 m/s / (14.5 sec) = 2 m/s^2 (approx), again consistent with our results.

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Self-critique (if necessary): ok

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Self-critique rating: ok

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Question: `q009. Describe in commonsense terms the motion of the object in this example if its initial velocity is indeed -8.9 m/s. Assume that the object starts at the crossroads between two roads running North and South, and East and West, respectively, and that the object ends up 80 meters North of the crossroads. In what direction does it start out, what happens to its speed, and how does it end up where it does?

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Your solution:

If the object is moving north or travelling up a slope as if on a graph, we could infer that the acceleration is decreasing. If it were moving south it would be travelling down the slope of a graph and would be accelerating more quickly.

confidence rating #$&*:

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Given Solution:

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Question: `q010. An object speeds up from 10 m/s to 20 m/s, accelerating uniformly and traveling 60 meters during this interval.

Specify which of the quantities v_0, v_f, aAve, `ds and `dt are given, and specify the value of each.

Specify which of the four equations of uniformly accelerated motion include the given three quantities. There is at least one such equation, and there might be two.

For each of the equations you specified, identify the quantity for which the value is not given. Then symbolically solve the equation for each of these quantities, showing the steps of your algebra.

Substitute the three given quantities into your solution, and simplify.

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Your solution:

Assuming the acceleration is uniform we can say that the initial velocity is 10m/s and final velocity is 20 m/s. This means our average velocity is 10m/s+20m/s / 2 = 15m/s. From our average velocity we can determine our time displacement with `dt = `ds (60m) / vAve (15m/s) = 4 sec. Once we have our time interval we can figure acceleration, which is the change in velocity (10m/s) / change in time (4 sec) = 2.5 m/s/s.

confidence rating #$&*:we had not assumed the initial and final velocities were 10m/s and 20 m/s, we would not have had sufficient data to solve for the remaining quantities. Im not sure exactly which of our 4 fundamental equations this employs, they are most likely algebraic rearrangements of vf=v0+a `dt which can be changed to find the acceleration, the change in time, or the initial or final velocities. In the latter case of finding either initial or final velocity we solve for `dt in the 1st fundamental equation to get our initial velocity with v0=2*`ds / `dt - vf. If we solve for `dt in the 2nd fundamental equation we end up with vf^2= v0^2 +2a * `ds we can achieve our final velocity. ????I am still refreshing myself on some of the algebraic concepts here, please confirm with me that my reasoning in this problem is correct or not????

@&
Your reasoning is fine, and your use of equations is progressing.

The quantities you are given in this equation are

v0, vf and `ds.

The equation vf = v0 + a `dt contains two of the given variables, but it also contains the two variables a and `dt, which are not given.

So this equation is of no direct use, but might become useful at a later stage of the problem.

The first equation

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

on the other hand contains all three of the given quantities, with only `dt unknown. So we can solve the equation for `dt, and use our solution to find the value of `dt.

This is actually more complicated than direct reasoning, but it illustrates the general scheme of using the equations, and it's necessary practice for situations in which direct reasoning doesn't suffice.

*@

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Question: `q011. An cart initially moving at 10 cm/s travels 40 cm while accelerating at 5 m/s^2. Using the equations of uniformly accelerated motion determine the time required and the the cart's final velocity.

Hint: You will need to start out with either the third or the fourth equation of uniformly accelerated motion. You are advised that it's easier to start out with the fourth equation.

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Your solution:

Using the 4th fundamental equation is preferred because it’s ready to have the units plugged in and simplified, the 3rd fundamental equation requires some algebra to get to the 4th equation. The equation is

Vf^2 = v0^2 + 2 (a * `ds)

= `sqrt [(10cm/s)^2 + 2 (5cm/s^2 * 40cm)]

= `sqrt [100cm^2/s^2 + 2 (200cm^2/s^2)]

= `sqrt [100cm^2/s^2 + 400cm^2/s^2]

= `sqrt [500cm^2/s^2]

fv = 22.4cm/s

From here we can find the average velocity by adding the initial and newly achieved final velocity and dividing by 2.

vAve= (10cm/s + 22.4 cm/s) / 2 = 16.2 cm/s

With the vAve in our possession we can now determine the time interval by dividing the change in position by the newly achieved average velocity.

`dt = 40cm / 16cm/s = 2.5 sec

confidence rating #$&*:

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This was the first one I was able to confidently do without referring to a solution. They’re much easier when worked out on paper.

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Question: `q012. An object starts at position x = +10 cm with a velocity of +5 cm/s, and accelerates uniformly, ending up at position x = -30 cm after a time interval of 8 seconds. What is its velocity at this point, and what was its acceleration during this interval?

Principles of Physics students should not spend over 5 minutes on this problem, General College Physics students should not spend over 10 minutes.

University Physics students are expected to be able to solve this problem, but if it hasn't been solved within 15 minutes, should submit their best thinking and await the instructor's notes.

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Your solution:

From the change in position divided by the change in time we observe our average velocity

vAve = (-20cm/s) / 8 sec = -2.5 sec

From the average velocity we can derive our final velocity, the only number that when added to 5cm and divided by 2 to produce -2.5 cm/s is -10cm.

The change in velocity is therefore -15 sec. We then divide the change in velocity by the change in time to get our average acceleration at -1.875 cm/s^2

@&
Very good reasoning. However the change from +10 cm/s to -30 cm/s is -40 cm/s, not -20 cm/s. No need to revise anything, but it would be easy to do so.
*@

confidence rating #$&*:

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The object ends up at position +80 meters, which is assumed to be 80 meters to the North. Its initial velocity is -8.9 m/s, with the - sign indicating that the initial velocity is in the direction opposite to the displacement of the object. So the object must start out moving to the South at 8.9 meters/second.

Its acceleration is +2 m/s^2, which is in the opposite direction to its initial velocity. This means that the velocity of the object changes by +2 m/s every second. After 1 second the velocity of the object will therefore be -8.9 m/s + 2 m/s = -6.9 m/s. After another second the velocity will be -6.9 m/s + 2 m/s = -4.9 m/s. After another second the velocity will be -2.9 m/s, after another -.9 m/s, and after another -.9 m/s + 2 m/s = +1.1 m/s. The speed of the object must therefore decrease, starting at 8.9 m/s (remember speed is always positive because speed doesn't have direction) and decreasing to 6.9 m/s, then 4.9 m/s, etc. until it reaches 0 for an instant, and then starts increasing again.

Since velocities after that instant become positive, the object will therefore start moving to the North immediately after coming to a stop, picking up speed at 2 m/s every second. This will continue until the object has attained a velocity of +20 meters/second and has displaced +80 meters from its initial position.{}{}It is important to understand that it is possible for velocity to be in one direction and acceleration in the other. In this case the initial velocity is negative while the acceleration is positive. If this continues long enough the velocity will reach zero, then will become positive.

STUDENT QUESTION

I understood the negative velocity but was unsure how to explain the rest. I am still rather confused by the last paragraph, especially where it says that it is possible for velocity to be in one direction and acceleration in the other.

INSTRUCTOR RESPONSE

If you speed up the acceleration is in the direction of motion.

If you slow down the acceleration is opposite the direction of motion.

To speed up a wagon you can get behind it and push in the direction of its motion, giving it an acceleration in its direction of motion.

To slow it down you can get in front of it and push it against its direction of motion (not advisable if it's a big wagon; think of stopping a child in a small wagon), giving it an acceleration in the direction opposite its motion.

STUDENT COMMENTS

Made relative sense, but still unsure in doubt to my answer that the object was increasing after it was moving in the

right direction. Being negative, would it have started towards the south at that acceleration, and moving north would it have

diminished its negativity? It seems this naturally.

INSTRUCTOR RESPONSE

Accelerating to the north (not 'moving' to the north; it does end up moving to the north, but the object starts out moving to the south), the speed in the southward direction would have diminished, as you say.

Eventually it comes to rest, just for an instant, somewhere south of its starting point. Then the northward acceleration will give it an increasing northward velocity.

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Self-critique (if necessary): ok

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Self-critique rating: ok

Questions, Problems and Exercises

You should answer the questions and work the problems as given below, as appropriate to your course. Your work should normally be handwritten, should include diagrams and sketches where appropriate, and should go into your Physics notebook (not into your lab notebook).

If the course is not specified for a problem, then students in all physics courses should do that problem.

Principles of Physics students need not do the questions or problems that are specified for General College Physics or University Physics.

General College Physics students need not do questions or problems specified for University Physics.

University Physics students should do all questions and problems.

Principles of Physics students may if they wish do some of the questions and problems specified for General College Physics, but this is neither expected nor required. These problems are accessible to Principles of Physics students, but are generally more challenging that what is expected of students in this course. (Some University Physics problems will also be accessible to Principles of Physics students, though some will not.)

General College Physics students who wish to do so are welcome to work some or all of the University Physics questions and problems, though this is neither expected nor required. Many of the University Physics questions and problems are more challenging than those expected of General College Physics students, and a number of the problems require the use of calculus, which is not expected of General College Physics students.

You are not expected to submit these questions and problems. It would take too much of your time to key in all the answers and solutions. You should solve these problems and answer these questions in your notebook, in a form you can later reference and, if you later desire, revise. The Query at the end of the assignment will ask you selected questions, which you will at that time be expected to answer based on the work you have done in your notebook.

Remember that you are always welcome to ask questions at any time. Any question about a problem should include a copy of the problem and a summary of what you do and do not understand about it.

Questions related to q_a_

1. Using the equations of uniformly accelerated motion, find the initial velocity of a coasting automobile which accelerates at a uniform rate of 2 m/s^2 for 12 seconds, ending up with a velocity of 30 m/s. Also determine how far the automobile coasts during this interval.

Then explain how to reason this problem out using the definitions of velocity and acceleration, without reference to the equations of uniformly accelerated motion.

2. Using the equations of uniformly accelerated motion, find the initial velocity of a coasting automobile which accelerates at a uniform rate of 2 m/s^2 as it travels 125 meters, ending up with a velocity of 30 m/s. Also determine how long it takes the automobile to coast this distance.

Using the initial velocity you found, the given final velocity, and the time interval you found, use direct reasoning, in terms of the definitions of velocity and acceleration, to find the acceleration and displacement. Verify that your results agree with the given acceleration and displacement.

3. Show how to solve the equation `ds = v0 `dt + .5 a `dt^2 for v0.

4. The fourth equation of uniformly accelerated motion is vf^2 = v0^2 + 2 a `ds. Suppose that a = .5 m/s^2, `ds = 9 m and v0 = 4 m/s. Solve for vf. (recall that there are two solutions to the equation x^2 = c; the solutions are x = sqrt(c) and x = - sqrt(c); for example if the equation is x^2 = 9 then both x = 3 and x = -3 are clearly solutions)

As you see, there are two solutions for vf, one with a positive value and one with a negative value.

Using the positive value of vf:

• Use the first equation of uniformly accelerated motion, along with given quantities, to find `dt.

• Use the second equation of uniformly accelerated motion, along with given quantities, to find `dt.

• Using your values of v0, vf and `dt, apply the definitions of average velocity and average acceleration to reason out the values of `ds and a, and verify that the values you get by reasoning agree with the values given in the problem.

Repeat the preceding, using the negative value of vf.

5. An automobile coasts at high speed down a hill, speeding up as it goes. It encounters significant air resistance, which causes it to speed up less quickly than it would in the absence of air resistance.

If the positive direction is down the hill, then

• Is the direction of the automobile's velocity positive or negative?

• Is the direction of the automobile's acceleration positive or negative?

• Is the direction of the air resistance positive or negative?

• Is the direction of the automobile's displacement, from start to finish, positive ornegative?

If the positive direction is up the hill, then

• Is the direction of the automobile's velocity positive or negative?

• Is the direction of the automobile's acceleration positive or negative?

• Is the direction of the air resistance positive or negative?

• Is the direction of the automobile's displacement, from start to finish, positive or negative?

Questions related to Class Notes

1. A ball rolls a variety of distances down an incline, from rest, and from a variety of starting positions. The corresponding intervals are timed using the TIMER program. All time intervals last at least 1 second but none exceeds 2 seconds. The resulting acceleration values range from 43.2 cm/s^2 to 49.6 cm/s^2.

• According to your experience, what is a reasonable percent uncertainty in measuring these time intervals?

• Assuming this percent uncertainty, are these results consistent with the hypothesis that acceleration on the incline is independent of position or velocity on the incline?

Questions/problems for General College Physics Students

1. My garden is 30 meters long and 15 meters wide. To protect against drought I want to build a pond containing a month's supply of water, equivalent to about 3 inches of rainfall. If the pond is 10 meters long and 6 meters wide, how deep does it have to be?

2. At 1200 liters/day per family how much would the level of a lake with surface area 50 km^2 fall in a year if supplying town of population 40000?

3. Estimate the number of gallons of gasoline used by all drivers in the United States in a month. Base your estimate on reasonable assumptions. State your assumptions and explain how they lead to your conclusion.

Questions/problems for University Physics Students

1. A sailor sails 2 km due east, then 3.5 km toward the southeast, then through an unknown displacement. He ends up 5.8 km to the east of his starting point.

• Find the magnitude and direction of the third leg of his path.

• Sketch a diagram and explain how your diagram shows qualitative agreement with your solution.

2. We have two vectors, one of magnitude 3.6 directed at an angle of +70 deg relative to the positive x axis, and another of magnitude 2.4 directed at +210 deg from the positive x axis.

• Find the scalar product of these vectors.

• Find the vector product of these vectors.

Self-critique (if necessary):

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Questions, Problems and Exercises

If the course is not specified for a problem, then students in all physics courses should do that problem.

Principles of Physics students need not do the questions or problems that are specified for General College Physics or University Physics.

General College Physics students need not do questions or problems specified for University Physics.

University Physics students should do all questions and problems.

Remember that you are always welcome to ask questions at any time. Any question about a problem should include a copy of the problem and a summary of what you do and do not understand about it.

Questions related to q_a_

Then explain how to reason this problem out using the definitions of velocity and acceleration, without reference to the equations of uniformly accelerated motion.

3. Show how to solve the equation `ds = v0 `dt + .5 a `dt^2 for v0.

As you see, there are two solutions for vf, one with a positive value and one with a negative value.

Using the positive value of vf:

• Use the first equation of uniformly accelerated motion, along with given quantities, to find `dt.

• Use the second equation of uniformly accelerated motion, along with given quantities, to find `dt.

Repeat the preceding, using the negative value of vf.

If the positive direction is down the hill, then

• Is the direction of the automobile's velocity positive or negative?

• Is the direction of the automobile's acceleration positive or negative?

• Is the direction of the air resistance positive or negative?

• Is the direction of the automobile's displacement, from start to finish, positive ornegative?

If the positive direction is up the hill, then

• Is the direction of the automobile's velocity positive or negative?

• Is the direction of the automobile's acceleration positive or negative?

• Is the direction of the air resistance positive or negative?

• Is the direction of the automobile's displacement, from start to finish, positive or negative?

Questions related to Class Notes

• According to your experience, what is a reasonable percent uncertainty in measuring these time intervals?

• Assuming this percent uncertainty, are these results consistent with the hypothesis that acceleration on the incline is independent of position or velocity on the incline?

Questions/problems for General College Physics Students

2. At 1200 liters/day per family how much would the level of a lake with surface area 50 km^2 fall in a year if supplying town of population 40000?

Questions/problems for University Physics Students

1. A sailor sails 2 km due east, then 3.5 km toward the southeast, then through an unknown displacement. He ends up 5.8 km to the east of his starting point.

• Find the magnitude and direction of the third leg of his path.

• Sketch a diagram and explain how your diagram shows qualitative agreement with your solution.

2. We have two vectors, one of magnitude 3.6 directed at an angle of +70 deg relative to the positive x axis, and another of magnitude 2.4 directed at +210 deg from the positive x axis.

• Find the scalar product of these vectors.

• Find the vector product of these vectors.

Self-critique (if necessary):

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Self-critique rating:

#*&!

Open QA 6

@& Very good reasoning. However the change from +10 cm/s to -30 cm/s is -40 cm/s, not -20 cm/s. No need to revise anything, but it would be easy to do so. *@

&#This looks good. See my notes. Let me know if you have any questions. &#

@&
Very good reasoning. However the change from +10 cm/s to -30 cm/s is -40 cm/s, not -20 cm/s. No need to revise anything, but it would be easy to do so.
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