qa 06

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course Phy 201

2:15 pm EST 6/18/14

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.

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:

a = (vf-v0)/'dt. (30-10m/s)/15s = 1.33m/s^2.

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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Question: `q002. It wasn't necessary to use a 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:

Acceleration is simply the change in velocity over time. Or, to put it another way, how fast is the velocity changing? How fast is it speeding up or slowing down? For

that, we know velocity increases 20m/s. Divide this by 15 to get 1.33m/s^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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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:

Step 1: 'ds = (vf+v0)/2 * 'dt

Step 2: Divide by 'dt on both sides for, 'ds/'dt = (vf + v0)/2

Step 3: Multiply by 2 on both sides for, 2 * 'ds/'dt = vf + v0

Step 4: Subtract vf from both sides for, v0 = 2'ds/'dt - vf.

Step 5: Plug in numbers, 2(80m/10s) - 6m/s = 10m/s

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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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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: The elements for vAve are given, 80m/10s = 8m/s. If vAve could also be initial plus final velocity divided by 2, what when added to 6m/s and divided by

2 equal 8m/s? Well, since 16/2 = 8, 10m/s plus 6m/s / 2 would equal 10m/s, giving a initial velocity of 10m/s.

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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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: First, subtract -1/2a'dt^2 from both sides to get, 'ds - 1/2a'dt^2 = v0t. Then divide both sides by 'dt to get v0. Then plug in the numbers 80 - (1/2 *

-2m/s^2 * 10s^2)/10s = v0. or 180/10 = 18m/s.

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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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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: The change in velocity equals acceleration multiplied by a given time interval. So, 'dv = -2m/s^2 * 10s = -20m/s. Add this to v0 and divide by 2 to get

vAve of 8m/s. 8m/s * 10s = 80m.

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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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:

Step 1: vf^2 = v0^2 + 2a*`ds

Step 2: vf^2 - 2a*'ds = v0^2

Step 3: 20m/s^2 - (2*2m/s^2*80m) = v0

400 - 320 = sq root of 80 = +/-8.9m/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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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: The average velocity of (20+8.9m/s)/2 = 14.45m/s, consistent with results above. 'dt = 'ds/vAve or 80m/14.45m/s = 5.5s. The change in velocity of

11.1m/s / 5.5 s equal about 2m/s^2, consistent with acceleration from above.

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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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: This is too complex without visuals. I don't know how you calculated acceleration.

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Given Solution:

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, expecially 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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If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following

problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

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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: v_0 is 10m/s. v_f is 20m/s. 'ds is 60m. 'ds = (v_f - v_0)/2 * 'dt. 'dt = 60m/15m/s = 4s. Before doing this, we find vAve by (10m/s + 20m/s) / 2 = 15m/s.

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Question: `q011. An cart initially moving at 10 cm/s travels 40 cm while accelerating at 5 cm/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: v_f^2 = v_0^2 + 2a*ds. Once I plug in the numbers and take the sq root v_f^2, I get 12.25m/s. vAve = (12.25 + 10m/s) / 2 = 11.13cm/s. 'dt = 'ds/vAve =

40cm/11.13cm/s = 3.6s.

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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: I don't know.

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You should submit the above questions, along with your answers and self-critiques. You may then begin work on the Questions, Problem and Exercises, as instructed

below.

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.

v0 = vf - (a * 'dt) = 30m/s - (2m/s^2 * 12s) = 6m/s

'ds = (v0 + vf)/2 * 'dt = (6m/s + 30m/s)/2 * 12s = 216m

You find initial velocity by taking the final velocity and subtracting the change in velocity throughout this time period multiplied by the time interval.

you find how far it traveled by finding how fast is was going on average and multiply by the during of time is was going at this average speed to find the distance it

traveled.

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.

Subtract 0.5a""dt^2 from both sides to get: ('ds - 0.5a""dt^2)/t = v0'dt.

Then divide by 'dt on both sides to get: v0 = 'ds - 0.5a'dt

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."