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course Phy201+L

006. Using equations with uniformly accelerated motion.

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Question: `q001. Note that there are 9 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:

To solve for a, you first subtract v0 from both sides resulting in the equation vf-v0= a * ‘dt. Then divide both sides by ‘dt, resulting in the final equation of (vf - v0)/ ‘dt=a.

Then, inserting our values, we get the equation:

(30 m/s - 10 m/s) / 15 sec = 1.33 m/s/s

a=1.33 m/s/s

Confidence rating: 3

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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):

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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:

You could see that you have a change in velocity over a time period, so seeing that initial information, you could determine that the difference in the velocities divided by the time interval would give you acceleration.

confidence rating #$&*:32; 3

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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:

The first step is to solve for v0. First we can multiply both sides by 2 * ‘dt, resulting in the equation 2 * ‘ds/ ‘dt = vf + v0. Then subtract both sides by vf resulting in the final equation of, 2 * ‘ds / ‘dt - vf = v0.

Then insert the given values to determine the initial velocity:

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

v0 = 10 m/s

confidence rating #$&*:32; 3

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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:

To obtain the average velocity we can divide our given 80 m by our given time period of 10 sec to give us the velocity quantity of 8 m/sec. Knowing that average velocity is a average between the initial velocity and the final velocity, we can see that the difference between the final velocity of 6 m/sec and the average velocity is 2, we can determine that the initial velocity was 10 m/s.

confidence rating #$&*:32; 3

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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:

The first step is to solve for v0. Our first step is to subtract .5a’dt^2, resulting in the equation ’ds - .5 a ‘dt^2= v0 ‘dt. We then divide both sides of the equation by ‘dt, giving us the final equation of (‘ds - .5 a ‘dt^2)/ ‘dt = v0.

Then inserting our values we can determine v0:

((80 m) - .5 (-2m/s/s) (10 sec)^2)/ (10 sec) = 18 m/s

v0= 18 m/s

confidence rating #$&*:32; 3

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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 of an object with an acceleration of -2 m/s/s over a span of 10 seconds would be:

-2 m/s/s * 10 sec = -20 m/s

The final velocity will therefore be:

18 m/s -20 m/s = -2 m/s

The average velocity will be:

(18 m/s - 2m/s)/2 = 8 m/s

An object moving with an average velocity of 8 m/s over a span of 10 seconds will travel 80 meters.

Confidence rating: 3

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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:

The first step is to solve the equation for v0. Our first step is to subtract - 2 a ‘ds from both sides of the equation to give us:

vf^2 - 2 a ‘ds = v0^2. We then take the square root of both sides to give us the final equation of v0= sqareroot(vf^2 - 2 a ‘ds). Then we insert our values to determine v0:

( 20 m/s) ^2 - 2 (2 m/s/s) (80 m) = 80 m/s

Then take the square root of 80 m/s. This gives us a value of +/- 8.9 m/s.

Our initial velocity (v0) is +/- 8.9 m/s.

confidence rating #$&*:32; 3

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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:

In the case of -8.9 m/s, we have to add the values (-8.9 m/s + 20 m/s) /2 = 5.5 m/s (average velocity)

The change in velocity will be the final velocity minus the initial velocity (20 m/s - (-8.9 m/s)) = 28.9 m/s.

Time to cover 80 meters :

80m/ 5.5 m/sec = 14.5 sec.

A velocity change of 28.9 m/s over a span of 14.5 sec (28.9 m/s / 14. 5 s) would give us an average acceleration of 2 m/s/s, as our results show.

Confidence rating: 3

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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:

The object will begin moving south at 8.9 m/s initially but it will slow down at a rate of 2m/s/s as our acceleration shows, until it reaches zero, when it begins to head in the positive direction (North).

confidence rating #$&*:32; 3

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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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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. vf = v0 + a * `dt

vf- a * ‘dt = v0

30 m/s - 2m/s/s *12 sec = 6 m/s

v0= 6 m/s

30 m/s - 6 m/s = 24 m/s (change in velocity)

24 m/s * 12 sec = 288 meters.

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.

vf^2 = v0^2 + 2 a `ds

vf^2 - 2 a ‘ds = v0^2

sq. rt.( vf^2 -2 a ‘ds) = v0

((30 m/s)^2 - 2 (2m/s/s)(125))sq.rt.= 20 m/s (v0)

(30 m/s + 20 m/s)/2 = 25 m/s

25 m/s /125 m = 0.2 meters.

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 - 0.5 a ‘dt ^2 from both sides giving the equation ‘ds- 0.5 a ‘dt ^2= v0 ‘dt. Then divide both sides by ‘dt giving the final equation of (‘ds -0.5 a ‘dt^2)/’dt = 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)

sq.rt. (v0^2 + 2 a ‘ds) = vf

((4)^2 + 2 (.5)(9)) sq.rt. = +/- 5 m/s (vf)

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?

• POSITIVE.

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

• POSITIVE.

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

• NEGATIVE.

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

• POSITIVE.

If the positive direction is up the hill, then

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

• Negative.

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

• Negative.

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

• Positive.

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

• 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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Self-critique rating:

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_

vf- a * ‘dt = v0

30 m/s - 2m/s/s *12 sec = 6 m/s

v0= 6 m/s

30 m/s - 6 m/s = 24 m/s (change in velocity)

24 m/s * 12 sec = 288 meters.

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

vf^2 = v0^2 + 2 a `ds

vf^2 - 2 a ‘ds = v0^2

sq. rt.( vf^2 -2 a ‘ds) = v0

((30 m/s)^2 - 2 (2m/s/s)(125))sq.rt.= 20 m/s (v0)

(30 m/s + 20 m/s)/2 = 25 m/s

25 m/s /125 m = 0.2 meters.

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

Subtract - 0.5 a ‘dt ^2 from both sides giving the equation ‘ds- 0.5 a ‘dt ^2= v0 ‘dt. Then divide both sides by ‘dt giving the final equation of (‘ds -0.5 a ‘dt^2)/’dt = v0

sq.rt. (v0^2 + 2 a ‘ds) = vf

((4)^2 + 2 (.5)(9)) sq.rt. = +/- 5 m/s (vf)

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?

• POSITIVE.

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

• POSITIVE.

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

• NEGATIVE.

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

• POSITIVE.

If the positive direction is up the hill, then

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

• Negative.

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

• Negative.

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

• Positive.

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

• 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.

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Question: Given uniform acceleration 2 m/s^2, displacement 125 meters and final velocity 30 m/s, a student finds that the initial velocity is zero and the time interval is 10 seconds. We wish to check to see if the student's results are consistent with the given information.

• Using the initial velocity and time interval obtained by the student, along with the 30 m/s final velocity, quickly reason out the acceleration and the displacement in terms of the definitions of average velocity and/or average acceleration and the assumption of uniform acceleration.

• State whether the student's solutions are consistent with the originally given information.

• Compare with your solution to this problem.

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Your solution:

Given the final velocity of 30 m/s and the initial velocity of 0 m/s, the average velocity is 15 m/s. Multiplying the 15 m/s average velocity by the 10 second time interval, we get a displacement of 150 meters. The change in velocity is 30 m/s. The average acceleration would be the final velocity of 30 m/s divided by the 10 second time interval, giving us an average acceleration of 3 m/s/s.

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Given Solution:

The student obtained initial velocity zero and time interval 10 seconds. Along with the given 30 m/s final velocity this would lead to the following conclusions:

• the average velocity is 15 m/s, which since acceleration is uniform is simply the average of the initial and final velocities

justification in terms of definitions: if acceleration is uniform, then the v vs t graph is linear so that the average velocity is equal to the average of initial and final velocities

• multiplying the 15 m/s average velocity by the 10 s time interval interval we get displacement 150 m

justification in terms of definitions: ave velocity is ave rate of change of position with respect to clock time, so vAve = `ds / `dt, from which it follows that `ds = vAve * `dt

• the change in velocity would be 30 m/s, since the velocity changes on this interval from 0 m/s to 30 m/s; so the acceleration would be aAve = `dv / `dt = 30 m/s / (10 s) = 3 m/s^2

justification in terms of definitions: ave acceleration is ave rate of change of velocity with respect to clock time, so aAve = `dv / `dt.

in this situation acceleration is uniform, so we can if we wish use just plain a instead of aAve

The student's solution is not consistent with the given information, which specified acceleration 2 m/s^2 and displacement 125 meters.

A solution to the problem:

Using the fourth equation of motion with the given information (`ds, a and vf) we have

vf^2 = v0^2 + 2 a `ds , which we solve for v0 to get

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

+- sqrt( (30 m/s)^2 - 2 * (2 m/s^2) * (125 m) ) =

+- sqrt( 900 m^2 / s^2 - 500 m^2 / s^2) =

+- sqrt( 400 m^2 / s^2) =

+- 20 m/s.

If vf = 20 m/s then we could directly reason out the rest (vAve would be 25 m/s, so it would take 5 s to go 125 m), or we could use the first or second equation of motion to find `dt

The first equation `ds = (vf + v0) / 2 * `dt gives us `dt = 2 * `ds / (vf + v0) = 2 * 125 m / (20 m/s + 30 m/s) = 5 s.

Alternatively the second equation vf = v0 + a `dt gives us `dt = (vf - v0) / a = (30 m/s - 20 m/s) / (2 m/s^2) = 10 m/s / (2 m/s^2) = 5 s.

If vf = -20 m/s then we could directly reason out the rest (vAve would be (-20 m/s + 30 m/s) / 2 = 5 m/s, so it would take 25 s to go 125 m), or we could use the first or second equation of motion to find `dt

The first equation `ds = (vf + v0) / 2 * `dt gives us `dt = 2 * `ds / (vf + v0) = 2 * 125 m / (-20 m/s + 30 m/s) = 25 s.

Alternatively the second equation vf = v0 + a `dt gives us `dt = (vf - v0) / a = (30 m/s - (-20 m/s) ) / (2 m/s^2) = 50 m/s / (2 m/s^2) = 25 s.

STUDENT COMMENT

Umm, evidently I did NOT understand the problem.. even looking back, I’m still not sure how everything in the question exactly relates to the answer. I understand the given answer and what it means, but the original questions are very confusing to me! INSTRUCTOR RESPONSE

The point is that the student's solutions are inconsistent. Using the initial velocity obtained by the student, the change in velocity would be 30 m/s and the acceleration would be 3 m/s^2. This of course contradicts the given acceleration, which was 2 m/s^2. So the student's solution contradicts the given information.

STUDENT COMMENT

I do not understand this problem at all. The information given all makes sense and I know that I am looking for some way to make it all fit together to verify the students results. I understand it a little better and understand the path I was supposed to take after reading through the explanation, but am going to have to work on problems like these. INSTRUCTOR RESPONSE

It would be really beneficial for you to go through the given solution one phrase at a time, and tell me exactly what you do and do not understand. For example in the first few lines:

The student obtained initial velocity zero and time interval 10 seconds. Along with the given 30 m/s final velocity this would lead to the following conclusions:

• the average velocity is 15 m/s, which since acceleration is uniform is simply the average of the initial and final velocities

**** do you understand how the given information leads to the conclusion that the average velocity is 15 m/s? ******** what do you and do you not understand about the meaning of the statement 'since acceleration is uniform is simply the average of the initial and final velocities'? **** the change in velocity would be 30 m/s, since the velocity changes on this interval from 0 m/s to 30 m/s **** what do you and do you not understand about the details of this statement and its overall meaning? **** so the acceleration would be aAve = `dv / `dt = 30 m/s / (10 s) = 3 m/s^2  **** what do you and do you not understand about the meaning of the statement acceleration is `dv / `dt? ****  **** what do you and do you not understand about why in this case `dv is 30 m/s and `dt is 10 s? **** **** what do you and do you not understand about the calculation 30 m/s / (10 s) = 3 m/s^2? **** You should deconstruct the entire solution, one phrase at a time, and tell me what you do and do not understand about each. With that information I can help you address the things you don't understand.

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Question:

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 air resistance positive or negative?

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?

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Your solution:

The direction of the velocity is positive.

The direction of the air resistance is negative.

The automobile’s velocity is negative.

The direction of the automobile’s acceleration is negative.

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Given Solution:

If you are riding in the car your perception is that the forward direction is the one in which you are moving. You can stick your hand out the window and feel that the air resistance is in the 'backward' direction. Since the automobile is speeding up its acceleration is in its direction of motion.

The velocity is down the hill.

Thus the direction of the velocity and acceleration are both down the hill, and the direction of the air resistance is up the hill.

Therefore

• If the direction down the hill is positive then the velocity is positive, the acceleration is positive and air resistance is negative.

• If the direction down the hill is negative then the velocity is negative, the acceleration is negative and air resistance is positive.

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Question: (gen and univ phy) At 1200 liters/day per family how much would level of 50 km^2 lake fall in a year if supplying town of population 40000

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Your solution:

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Given Solution:

You should know how the milliliter, liter and cubic meter, three common measure of volume, are related:

• a milliliter is the volume of a cube 1 cm on a side

• a liter is the volume of a cube 10 cm on a side

• a cubic meter is the volume of a cube 1 meter on a side

so that

• 1 liter = (10 cm)^3 = 1000 cm^3 or 1000 milliliters

• 1 cubic meter = (100 cm)^2 = 1 000 000 milliliters

• 1 cubic meter = 1 000 000 milliliters / (1000 milliliters / liter) = 1000 liters

You should also understand the following images, which will allow you to visualize and thereby reason out these and similar relationships

• It takes 10 sides of length 10 cm to make a 1 meter side, so to fill a cubic meter with 10 cm cubes we would need 10 layers, each with 10 rows and each row with 10 cubes, for a total of 10 * 10 * 10 cubes. A cubic meter is therefore 10 * 10 * 10 liters, or 1000 liters.

• It takes 10 sides of length 1 cm to make a 10 cm side, so to fill a liter (a 10 cm cube) with 1 cm cubes we would need 10 layers, each with 10 rows and each row with 10 cubes, for a total of 10 * 10 * 10 cubes. A liter is therefore 10 * 10 * 10 milliliters, or 1000 milliliters.

It is also helpful to visualize the relationship between a cubic meter and a cubic kilometer:

• A kilometer is 1000 meters.

• A cubic kilometer is therefore (1000 meters) ^ 3 = 1 000 000 000 m^3, or 10^9 m^3, or a billion m^3.

• A cubic kilometer is visualized as 1000 layers each with 1000 rows each made up of 1000 one-meter cubes, for a total of 1000 * 1000 * 1000 = 1 000 000 000 one-meter cubes.

Finally you should understand what a cylinder is and how we find its area.

• The volume of a cylinder is equal to the area of its cross-section multiplied its altitude, as you saw in the q_a_initial_problems.

• The links below explain prisms and cylinders, and their volumes, in elementary terms:

http://www.mathsisfun.com/geometry/prisms.html

http://www.mathsisfun.com/geometry/cylinder.html

The solution:

A liter is 1/1000 of a cubic meter. It can be thought of as a cube 10 cm on a side. To fill a 1-meter cube t would take 10 rows with 10 cubes in each row to make a single layer 10 cm high, and 10 layers to fill the cube--i.e., 10 * 10 * 10 = 1000 one-liter cubes fill a 1-meter cube.

A km is 1000 meters so a km^2 is (1000 m)^2 = 10^6 m^2.

1200 liters / day per family, with 40,000 people at 4 persons / family, implies 10,000 families each using 1200 * .001 m^3 = 1.2 m^3 per day.

• Total usage would be 10,000 families * 1.2 m^3 / day / family * 365 days / year = 4.3 * 10^6 m^3 / year.

Visualize the surface of the lake as the base of a large large irregular cylinder. The volume of a cylinder is the product of the area of its base and its altitude.

• The volume of water corresponding to a depth change `dy is therefore `dy * A, where A is the area of the lake.

• It might be helpful to think of a layer of ice several centimeters thick on top of the lake. The cross-sections of this layer all have very nearly the same size and shape, so it can be viewed as a cylinder with cross-sectional area equal to the area of the lake, and a thickness of several centimeters.

The area of the lake is 50 km^2 * 10^6 m^2 / km^2 = 5 * 10^7 m^2.

`dy * A = Volume so

• `dy = Volume / A

= 4.3 * 10^6 m^3 / (5 * 10^7 m^2) = .086 m or 8.6 cm.

This estimate is based on 4 people per family. A different assumption would change this estimate.

STUDENT QUESTION: 40000 people in town divided by average household of 4 = 10,000 families 10,000 families * 1200 liters /day = 12000000 liters used per day * 365 days in a year = 4380000000 liter used. Here is where I get confused changing from liters to level of 50 km^2 to subtract.

INSTRUCTOR RESPONSE: If you multiply the area of the lake by the change in depth you get the volume of water used. You know the area of the lake and the volume of the water used, from which you can find the change in depth.

Of course you need to do the appropriate conversions of units. Remember that a liter is the volume of a cube 10 cm on a side, so it would take 10 rows of 10 such cubes to make one layer, then 10 layers, to fill a cube 1 meter = 100 cm on a side.

You should also see that a km^2 could be a square 1 km on a side, which would be 1000 meters on a side, to cover which would require 1000 rows of 1000 1-meters squares.

If you end of having trouble with the units or anything else please ask some specific questions and I will try to help you clarify the situation.

ANOTHER INSTRUCTOR COMMENT:

The water used can be thought of as having been spread out in a thin layer on top of that lake. That thin layer forms a cylinder whose cross-section is the surface of the lake and whose altitude is the change in the water level. The volume of that cylinder is equal to the volume of the water used by the family in a year. See if you can solve the problem from this model.

COMMON ERROR: Area is 50 km^2 * 1000 m^2 / km^2

INSTRUCTOR COMMENT:

Two things to remember: You can't cover a 1000 m x 1000 m square with 1000 1-meter squares, which would only be enough to make 1 row of 1000 squares, not 1000 rows of 1000 squares.

1 km^2 = 1000 m * 1000 m = 1,000,000 m^2. **

STUDENT QUESTION

Can you explain this part of the problem:The area of the lake is 50 km^2 * 10^6 m^2 / km^2 = 5 * 10^7 m^2.Where did 10^6 m^2 / km^2 come from? I think I understand the rest of the problem. INSTRUCTOR RESPONSE

1 km = 1000 m, or 10^3 m. So (1 km)^2 = (10^3 m)^2, or1 km^2 = 10^6 m^2. Thus 10^6 m^2 / (1 km^2) is a division of a quantity by an equal quantity. It follows that 10^6 m^2 / km^2 = 1 So the product 50 km^2 * 10^6 m^2 / km^2 is just 50 km^2 * 1, or 50 km^2. Of course 50 km^2 * 10^6 m^2 / km^2 is also equal to 5 * 10^7 m^2. Thus 50 km^2 = 5 * 10^7 m^2.

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Question: univ 1.74 (11th edition 1.70) univ sailor 2 km east, 3.5 km SE, then unknown, ends up 5.8 km east find magnitude and direction of 3d leg, explain how diagram shows qualitative agreement

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Your solution:

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Given Solution:

`a** Letting the three vectors be A, B and C (C unknown) and the x axis point east we have A at 0 degrees, B at 315 degrees and the resultant at 0 degrees.

Ax = 2, Ay = 0 (A is toward the East, along the x axis).

Bx = B cos(315 deg) = 2.47, By = B sin(315 deg) = -2.47.

Rx = 5.8, Ry = 0.

Since Ax + Bx + Cx = Rx we have Cx = Rx - Ax - Bx = 5.8 - 2 - 2.47 = 1.33.

Since Ay + By + Cy = Ry we have Cy = Ry - Ay - By = 0 - 0 - (-2.47) = 2.47.

C has magnitude sqrt( Cx^2 + Cy^2) = sqrt ( 1.33^2 + 2.47^2) = sqrt(7.9) = 2.8, representing 2.8 km.

C has angle arctan(Cy / Cx) = arctan(2.47 / 1.33) = 61 deg. **

STUDENT QUESTION

Why is it 315 deg and not 45? Would you write it 315 deg northwest of east since it goes in that direction?

INSTRUCTOR RESPONSE

With the x and y axes in standard position, with the x axis pointing east and the y axis north, the southeasterly direction lies in the fourth quadrant, at 315 degrees as measured counterclockwise from the positive x axis. If you measure your vectors from anywhere else you can't use the simple relationships Ax = A cos(theta) and Ay = A sin(theta). With this convention you can, and the signs of the trig functions automatically take care of the + and - signs of the components. You should of course also be able to use triangle trigonometry, but the circular trigonometry of this solution will be used in most of the solutions you will see in the queries and qa's.

STUDENT COMMENT/QUESTION:

I had this totally wrong. I am sooooo confused as the student above was why the B vector would be 315 degrees? How come it is in the 4th quadrant? I put my vector going in the SE direction. Why is this incorrect? INSTRUCTOR RESPONSE

You could orient your x-y coordinate system in any way you wish, as long as the positive y axis is at 90 degrees counterclockwise relative to the positive x axis. You could orient the system so that the easterly direction is the y direction, which would be consistent with your 90 degree angle for the 2 km vector. If so, the x direction would have to be toward the south. This would put the southeasterly direction in the first quadrant. The southeasterly direction would be at 45 degrees with this orientation. The given solution assumes the x axis to be pointing east, so the y axis points north, and the southeasterly direction is 'below' the x axis, in the 4th quadrant, at 315 degrees.

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Question: `q**** query univ 1.86 (11th edition 1.82) (1.66 10th edition) vectors 3.6 at 70 deg, 2.4 at 210 deg find scalar and vector product

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Given Solution:

`a** GOOD STUDENT SOLUTION WITH INSTRUCTOR COMMENT:

A * B = |A||B| cos(theta) = AxBx + Ay By + AzBz

3.6 * 2.4 * cos (140 deg) = -6.62

To check for consistency we can calculate the components of A and B:

Ax = 3.6 * cos(70 deg) = 1.23

Ay = 3.6 * sin(70 deg) = 3.38

Bx = 2.4 * cos (210 deg) = -2.08

By = 2.4 * sin(210 deg) = -1.2

dot product = 1.23 *-2.08 + 3.38 * -1.2 = -2.5584 + -4.056 = -6.61. Close enough.

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Cross product:

| A X B | = |A| |B| sin (theta) = 3.6 * 2.4 * sin(140) = 5.554.

Finding the components we have

(Ay * Bz - Az * By) i + (Az * Bx - Ax + Bz) j + (Ax * By - Ay * Bx) k =

((3.38*0)-(0*-1.2)] i + [(0 * -2.08) - (1.23 * 0)] j + [(1.23*-1.2)-(3.38 * -2.08)] k =

0 i + 0 j + 5.55 k,

or just 5.55 k, along the positive z axis ('upward' from the plane).

INSTRUCTOR COMMENT: Your conclusion is correct. The cross product must be at a right angle to the two vectors. To tell whether the vector is 'up' or 'down' you can use your result, 5.55 k, to see that it's a positive multiple of k and therefore upward.

The other way is to use the right-hand rule. You place the fingers of your right hand along the first vector and position your hand so that it would 'turn' the first vector in the direction of the second. That will have your thumb pointing upward, in agreement with your calculated result, which showed the cross product as being in the upward direction. **

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