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course Phy 121

9/20 3:02pm

008. Using the Acceleration of Gravity; summarizing the analysis of uniformly accelerated motion

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Question: `q001. There are 5 questions in this set.

The accepted value of the acceleration of gravity is approximately 980 cm/s^2 or 9.8 m/s^2. This will be the acceleration, accurate at most places within 1 cm/s^2, of any object which falls freely (i.e., without the interference of any other force), near the surface of the Earth.

If you were to step off of a table and were to fall 1 meter to the ground, without hitting anything in between, your motion would very nearly approximate that of a freely falling object.

How fast would you be traveling when you first reached the ground?

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Your solution: We know a, `ds and v0 but I don't think I can reason a flow diagram from here. I'll have to use an equation. vf^2 = v0^2 + 2 a `ds

vf^2 =0^2 + a 9.8m/s^2 *1m

vf^2 = 19.6m^2/s^2 Take `sqrt of each side

vf = 4.4m/s (approximately)

confidence rating #$&*:

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Given Solution:

You would have an initial vertical velocity of 0, and would accelerate at 9.8 m/s^2 in the same direction as your 1 meter vertical displacement.

You would also have a slight horizontal velocity (you don't step off of a table without moving a bit in the horizontal direction, and you would very likely maintain a small horizontal velocity as you fell), but this would have no effect on your vertical motion.

So your vertical velocity is a uniform acceleration with v0 = 0, `ds = 1 meter and a = 9.8 m/s^2. The equation vf^2 = v0^2 + 2 a `ds contains the three known variables and can therefore be used to find the desired final velocity. We obtain

vf = +- `sqrt( v0^2 + 2 a `ds) = +- `sqrt ( 0^2 + 2 * 9.8 m/s^2 * 1 m)= +- `sqrt ( 19.6 m^2 / s^2) = +- 4.4 m/s, approx.

Since the acceleration and displacement were in the direction chosen as positive, we conclude that the final velocity will be in the same direction and we choose the solution vf = +4.4 m/s.

STUDENT QUESTION

????how important is the plus or minus before a square root or an approximation of a square root and what is its significance????

INSTRUCTOR RESPONSE:

In some situations the + solution will be the one that fits the situation, while in others it will be the -, and in others both the + and - solutions can be valid.

An example:

If a ball is thrown into the air, rising say 20 meters before falling back, then it will be at the 15 meter height twice, once while rising and once while falling. At one of those times the velocity will be positive, at the other the velocity will be negative. So both the + and the - solutions are valid.

On the other hand you might want the velocity when the ball hits the ground. Only for a downward velocity will be valid for this question. Your equations will give you the + and the - solution, and you will have to determine which applies (i.e., which is downward, given your choice of the positive direction).

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Self-critique (if necessary):I should start using + and - going forward.

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Self-critique rating:3

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Question: `q002. If you jump vertically upward, leaving the ground with a vertical velocity of 3 m/s, how high will you be at the highest point of your jump?

Note that as soon as you leave the ground, you are under the influence of only the gravitational force. All the forces that you exerted with your legs and other parts of your body to attain the 3 m/s velocity have done their work and are no longer acting on you. All you have to show for it is that 3 m/s velocity. So as soon as you leave the ground, you begin experiencing an acceleration of 9.8 m/s^2 in the downward direction.

Now again, how high will you be at the highest point of your jump?

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Your solution:

The high point of the jump should be when your velocity is 0m/s. For this interval that is the terminal velocity. We eventually need to find the `ds for this interval. So let's make a flow diagram and see where that takes us.

First line: v0 (3m/s) vf (0m/s) a (-9.8m/s^2)

Second Line: vAVe with lines back to v0 and vf. `dv with lines back to v0 and vf.

Third line: `dt with lines back to a and `dv

Fourth line: `ds with lines back to vAve and `dt

Our `dv is vf - v0 = -3m/s. Our vAve is 3m/s/2 or 1.5m/s

our ``dt = `dv/a = (-3m/s) / -9.8m/s^2 = .3s (approx)

Lastly we find our `ds which is vAve* `dt

`ds = 1.5m/s * .3 s = .45meters or approx .5meters.

confidence rating #$&*:

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Given Solution:

From the instant the leave the ground until the instant you reach your highest point, you have an acceleration of 9.8 m/s^2 in the downward direction.

Since you are jumping upward, and since we can take our choice of whether upward or downward is the positive direction, we choose the upward direction as positive. You might have chosen the downward direction, and we will see in a moment how you should have proceeded after doing so.

For now, using the upward direction as positive, we see that you have an initial velocity of v0 = + 3 m/s and an acceleration of a = -9.8 m/s^2. In order to use any of the equations of motion, each of which involves four variables, you should have the values of three variables. So far you only have two, v0 and a. {}What other variable might you know? If you think about it, you will notice that when objects tossed in the air reach their highest point they stop for an instant before falling back down. That is precisely what will happen to you.

At the highest point your velocity will be 0. Since the highest point is the last point we are considering, we see that for your motion from the ground to the highest point, vf = 0. Therefore we are modeling a uniform acceleration situation with

v0 = +3 m/s, a = -9.8 m/s^2 and vf = 0.

We wish to find the displacement `ds. Unfortunately none of the equations of uniformly accelerated motion contain the four variables v0, a, vf and `ds.

This situation can be easily reasoned out from an understanding of the basic quantities. We can find the change in velocity to be -3 meters/second; since the acceleration is equal to the change in velocity divided by the time interval we quickly determine that the time interval is equal to the change in velocity divided by the acceleration, which is `dt = -3 m/s / (-9.8 m/s^2) = .3 sec, approx.; then we multiply the .3 second time interval by the 1.5 m/s average velocity to obtain `ds = .45 meters.

However if we wish to use the equations, we can begin with the equation vf = v0 + a `dt and solve to find

`dt = (vf - v0) / a = (0 - 3 m/s) / (-9.8 m/s^2) = .3 sec.

We can then use the equation

`ds = (vf + v0) / 2 * `dt = (3 m/s + 0 m/s) / 2 * .3 sec = .45 m.

This solution closely parallels and is completely equivalent to the direct reasoning process, and shows that and initial velocity of 3 meters/second should carry a jumper to a vertical height of .45 meters, approximately 18 inches. This is a fairly average vertical jump.

If the negative direction had been chosen as positive then we would have a = +9.8 m/s^2, v0 = -3 m/s^2 (v0 is be in the direction opposite the acceleration so if acceleration is positive then initial velocity is negative) and again vf = 0 m/s (0 m/s is the same whether going up or down). The steps of the solution will be the same and the same result will be obtained, except that `ds will be -.45 m--a negative displacement, but where the positive direction is down. That is we move .45 m in the direction opposite to positive, meaning we move .45 meters upward.

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Self-critique (if necessary):OK - I was initially skeptical as to the value of flow diagrams, but now I think they are very, very useful.

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Self-critique rating:OK

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Question: `q003. If you roll a ball along a horizontal table so that it rolls off the edge of the table at a velocity of 3 m/s, the ball will continue traveling in the horizontal direction without changing its velocity appreciably, and at the same time will fall to the floor in the same time as it would had it been simply dropped from the edge of the table.

If the vertical distance from the edge of the table to the floor is .9 meters, then how far will the ball travel in the horizontal direction as it falls?

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Your solution: The ball will continue to move horizantly in the positive direction until it hits the ground so its a bit like a race of the horizantal versus vertical. ( I seem to remember some ballistics problems like this from high school physics)

We will need to figure out how long it will take for the ball to fall the .9m to the ground. Then we'll use that `dt to determine how far the ball travelled horizantly.

We first approach this problem like the first question, how long does it take to fall from the table, in this case .9 meters?

We know a, `ds and v0 so likw in question #1 we will use an equation. vf^2 = v0^2 + 2 a `ds

vf^2 =0^2 + a 9.8m/s^2 *.9m

vf^2 = 17.6m^2/s^2 Take `sqrt of each side

vf = 4.2m/s (approximately)

We can figure vAVe = (vf + v0 )/2= (4.2m/s +0m/s)/2 = 2.1m/s

vAve = `ds/`dt

2.1m/s = .9m/`dt

`dt2.1m/s = .9m

`dt = .9m/2.1m/s = .43s (approximately)

Now we know the ball travelled for .42 seconds before it hit the ground. Tavelling at 3m/s *.42 s = 1.3 m (approx)

confidence rating #$&*:

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Given Solution:

A ball dropped from rest at a height of .9 meters will fall to the ground with a uniform vertical acceleration of 9.8 m/s^2 downward. Selecting the downward direction as positive we have

`ds = .9 meters, a = 9.8 m/s^2 and v0 = 0.

Using the equation `ds = v0 `dt + .5 a `dt^2 we see that v0 = 0 simplifies the equation to `ds = .5 a `dt^2, so

`dt = `sqrt( 2 `ds / a) = `sqrt(2 * .9 m / (9.8 m/s^2) ) = .42 sec, approx..

Since the ball rolls off the edge of the table with only a horizontal velocity, its initial vertical velocity is still zero and it still falls to the floor in .42 seconds. Since its horizontal velocity remains at 3 m/s, it travels through a displacement of 3 m/s * .42 sec = 1.26 meters in this time.

STUDENT COMMENT:

I don’t see why you don’t use the formula that I used in my answer.

INSTRUCTOR RESPONSE:

In your solution you chose to solve the third equation

ds = v0 * dt + 0.5 * a * dt^2

for `dt. Since v0 = 0 for the vertical motion, the equation simplifies and is easy to solve for `dt.

However I avoid using the third equation to solve for `dt because it is quadratic in `dt, and is therefore very confusing to most students. It is less confusing to use the fourth equation to find vf, after which we can easily reason out `dt.

Of course in the present case v0 = 0 and the equation becomes every bit as simple as the fourth equation; in fact when v0 = 0 it's simpler to use the equation you used. However most students have problems with special conditions and special cases, so I choose not to confuse the issue, and consistently use the fourth equation in my solutions.

My convention of using the fourth equation in this situation does no harm to students who understand how to solve the quadratic, and who know how consider and apply special conditions. You and other students who are sufficiently comfortable with the mathematics should always use the most appropriate option.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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.

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Question: `q004. A ball starts out with a horizontal velocity of 4 m/s and vertical velocity 0 m/s. It falls freely for 2 seconds.

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Your solution:

The horizantal `ds is easy to figure.

Travelling at 4m/s for 2 seonds the ball travels 8 m horizantally.

Knows the acceleration of gravity we can reason out the ball's vf at 2 seconds. Starting from 0 after 1 second it is travelling 9.8m/s. after two seconds it is traveling 19.6m/s.

vAVe = (19.6m/s +0)/2 = 9.8m/s travelling at this velocity for two seconds, the ball falls 19.6 meters vertically. If the direction of motion is downward this would be a positive 19.6 meters.

confidence rating #$&*:

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Question: `q005. A ball starts out with a horizontal velocity of 4 m/s and vertical velocity 0 m/s. It falls freely to the ground 20 meters below.

During its fall how far does it travel in the vertical direction, and how far in the horizontal direction?

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Your solution:

Unless something stops it, it will fall 20 meters in the vertical direction. We need to figure out how long this takes in order to figure out the horizantal `ds

This again is similar to the very first problem.

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

vf^2 =0^2 + a 9.8m/s^2 *20m

vf^2 = 392m^2/s^2 Take `sqrt of each side

vf = 19.8m/s (approximately)

vAve = (v0+vf)/2 = 19.8m/s/2 = 9.9m/s (approx)

`dt = `ds/vAve = 20m/9.9m/s = 2.0s (approx)

Since it falls for approx 2 seconds if it's velocity is 4m/s it travels approximately 8m horizantally.

confidence rating #$&*:

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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. If you fall freely, starting from rest, for 3 seconds, how far do you fall and how fast are you going at that instant?

2. If you take a running start and jump off the edge of a platform, moving at 5 meters / second in the horizontal direction, then

How far will you have fallen, in the vertical direction, after 1 second?

How far will you have traveled in the horizontal direction by that time, assuming that your horizontal speed doesn't change?

3. How long would it take to fall through a vertical distance of 20 meters, starting from rest?

In the situation of the preceding problem, how far would you travel in the horizontal direction while falling 20 meters in the vertical direction?

Questions related to text

1. How are the equations of uniformly accelerated motion expressed in your text?

These equations are expressed in terms of a specific interval on the webpage for your course, as

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

vf = v0 + a `dt

`ds = v0 `dt + .5 a `dt^2

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

How do the symbols in your text correspond to the symbols used here?

How do the equations in your text differ from the equations as given here?

Questions/problems for Principles of Physics Students

Before you answer the questions given here, think through the following:

What is the interpretation of the average slope of a velocity vs. clock time graph for a given interval?

What is the interpretation of the area beneath a velocity vs. clock time graph for a given interval?

What is the interpretation of the average slope of a position vs. clock time graph?

1. If the graph of position vs. clock time is increasing at a decreasing rate, what can be said about the graph of velocity vs. clock time?

2. If the graph of velocity vs. clock time is increasing, what can be said about the graph of position vs. clock time?

3. If four points of a position vs. clock time graph are (3 sec, 8 meters), (7 sec, 16 meters), (10 sec, 19 meters) and (12 sec, 20 meters), then:

What is the average velocity on each of the three intervals?

Does the velocity appear to be increasing or decreasing?

Do you expect that the velocity vs. clock time graph is increasing at an increasing rate, increasing at a decreasing rate, increasing at a constant rate, decreasing at an increasing rate, decreasing at a decreasing rate or decreasing at a constant rate, and why?

4. If four points of a velocity vs. clock time graph are (3 meters/sec, 8 meters), (7 meters/sec, 16 meters), (10 meters/sec, 19 meters) and (12 sec, 20 meters / sec), then:

What is the average acceleration on each of the two intervals?

Is the average acceleration increasing or decreasing?

Approximately how far does the object move on each interval? (General College Physics and University Physics students in particular: Do you think your estimates of the distances are overestimates or underestimates?)

Do you expect that the position vs. clock time graph is increasing at an increasing rate, increasing at a decreasing rate, increasing at a constant rate, decreasing at an increasing rate, decreasing at a decreasing rate or decreasing at a constant rate, and why?

Questions/problems for General College Physics Students

1. If the velocity of a falling object is given by the velocity function v(t) = 10 m/s^2 * t - 5 m/s, then

Find the velocities at t = 1, 3 and 5 seconds.

Sketch a velocity vs. clock time graph, showing and labeling the three corresponding points.

Estimate the displacement and acceleration on each of the two intervals.

Assuming that the t = 1 sec position is zero, describe the position vs. clock time graph, the velocity vs. clock time graph and the acceleration vs. clock time graph for the motion between t = 1 and t = 5 seconds.

Questions/problems for University Physics Students

1. Answer the preceding question, but this time based on position and acceleration functions obtained using the methods of calculus.

Self-critique (if necessary):

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Self-critique rating: