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course Phy 121

9/13 2:00pm

005. Uniformly Accelerated Motion

Preliminary notes:

On any interval there are seven essential quantities in terms of which we analyze the motion of a nonrotating object:

the time interval `dt between the beginning and the end of the interval

the displacement `ds of the object during the interval

the initial velocity v0, the velocity at the beginning of the interval

the final velocity vf, the velocity at the end of the interval

the average velocity vAve of the object during the interval

the change `dv in the velocity of the object during the interval

the average acceleration a_Ave of the object during the interval

You should remember these symbols and their meanings. You will be using them repeatedly, and you will soon get used to them.

You should at any time be able to list these seven quantities and explain the meaning of each.

In any question or problem that involves motion, you should identify the interval of interest, think about what each of these quantities means for the object, and identify which quantities can be directly determined from the given information.

You will of course improve your understanding and appreciation of these quantities as you work through the qa and the associated questions and problems.

Note also that `dt = t_f - t_0, where t_f represents the final clock time and t_0 the initial clock time on the interval, and that `ds = s_f - s_0, where s_f represents the final position and t_0 the initial position of the object on the interval.

Further discussion of symbols (you can just scan this for the moment, then refer to it when and if you later run into confusion with notation)

the symbol x is often used instead of s for the position of an object moving along a straight line, so that `dx might be used instead of `ds, where `dx = x_f - x_0

some authors use either s or x, rather that s_f or x_f, for the quantity that would represent final position on the interval; in particular the quantity we express as `dx might be represented by x - x_0, rather than x_f - x_0

some authors use t instead of `dt; there are good reasons for doing so but at this point in the course it is important to distinguish between clock time t and time interval `dt; this distinction tends to be lost if we allow t to represent a time interval

the quantity we refer to as `dt is often referred to as 'elapsed time', to distinguish it from 'clock time'; once more we choose here to use different symbols to avoid confusion at this critical point in the course)

If the acceleration of an object is uniform, then the following statements apply. These are important statements. You will need to answer a number of questions and solve a number of problems in order to 'internalize' their meanings and their important. Until you do, you should always have them handy for reference. It is recommended that you write a brief version of each statement in your notebook for easy reference:

1. A graph of velocity vs. clock time forms a straight line, either level or increasing at a constant rate or decreasing at a constant rate.

2. The average velocity of the object over any time interval is equal to the average of its velocity at the beginning of the time interval (called its initial velocity) and its velocity at the end of the time interval (called its final velocity).

3. The velocity of the object changes at a constant rate (this third statement being obvious since the rate at which the velocity changes is the acceleration, which is assumed here to be constant).

4. The acceleration of the object at every instant is equal to the average acceleration of the object.

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Question: `q001. Note that there are 13 questions in this assignment.

Suppose that an object increases its velocity at a uniform rate, from an initial velocity of 5 m/s to a final velocity of 25 m/s during a time interval of 4 seconds.

By how much does the velocity of the object change?

What is the average acceleration of the object?

What is the average velocity of the object?

(keep your notes on this problem, which is continued through next few questions)

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Your solution:

Velocity change is determined using the formula `dv = vf-v0 = 25m/s - 5 m/s = 20 m/s

aAve = `dv/dt = (20/s)/4 sec = 5 m/s^2

vAve = (v0 + vf)/2 = (5m/s + 25m/s)/2 = (30m/s)/2 = 15m/s

confidence rating #$&*:

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Given Solution:

The velocity of the object changes from 5 meters/second to 25 meters/second so the change in velocity is 20 meters/second. The average acceleration is therefore (20 meters/second) / (4 seconds) = 5 m / s^2. The average velocity of the object is the average of its initial and final velocities, as asserted above, and is therefore equal to (5 meters/second + 25 meters/second) / 2 = 15 meters/second (note that two numbers are averaged by adding them and dividing by 2).

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: `q002. How far does the object of the preceding problem travel in the 4 seconds?

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Your solution:

`ds = vAVE * `dt = 15m/s * 4 s = 60 m

confidence rating #$&*:

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Given Solution:

The displacement `ds of the object is the product vAve `dt of its average velocity and the time interval, so this object travels 15 m/s * 4 s = 60 meters during the 4-second time interval.

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Self-critique (if necessary):Ok

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Self-critique rating:ok

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Question: `q003. Explain in commonsense terms how we determine the acceleration and distance traveled if we know the initial velocity v0, and final velocity vf and the time interval `dt.

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Your solution:

To determine acceleration we take the change in velocity and divide it by the time interval. We know the time interval (or change in clock time) so we need to figure out the change in velocity. For this we subtract the initial velocity from the final velocity. Lastly we take this change in velocity and divide it by the time interval.

To determine the distance travelled we need to figure out the average velocity and multiply it by the time we travelled. To find the average velocity we add the initial velocity and the final velocity and divide this sum by two. This is the average velocity. When we multiply this by the time spent travelling, we have the distance traveled.

confidence rating #$&*:

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Given Solution:

In commonsense terms, we find the change in velocity since we know the initial and final velocities, and we know the time interval, so we can easily calculate the acceleration. Again since we know initial and final velocities we can easily calculate the average velocity, and since we know the time interval we can now determine the distance traveled.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: `q004. Symbolize the situation by first giving the expression for the acceleration in terms of v0, vf and `dt, then by giving the expression for vAve in terms of v0 and vf, and finally by giving the expression for the displacement in terms of v0, vf and `dt.

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Your solution:

aAve = (vf-v0)/`dt

vAve =(v0 +vf)/2

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

confidence rating #$&*:

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Given Solution:

The acceleration is equal to the change in velocity divided by the time interval; since the change in velocity is vf - v0 we see that the acceleration is a = ( vf - v0 ) / `dt.

The average velocity is the average of the initial and final velocities, which is expressed as (vf + v0) / 2.

When this average velocity is multiplied by `dt we get the displacement, which is `ds = (v0 + vf) / 2 * `dt.

STUDENT SOLUTION (mostly but not completely correct)

vAve = (vf + v0) / 2

aAve = (vf-v0) / dt

displacement = (vf + v0)/dt

INSTRUCTOR RESPONSE

Displacement = (vf + v0)/dt is clearly not correct, since greater `dt implies greater displacement. Dividing by `dt would give you a smaller result for larger `dt.

From the definition vAve = `ds / `dt, so the displacement must be `ds = vAve * `dt. Using your correct expression for vAve you get the correct expression for `ds.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: `q006. This situation is identical to the previous, and the conditions implied by uniformly accelerated motion are repeated here for your review: If the acceleration of an object is uniform, then the following statements apply:

1. A graph of velocity vs. clock time forms a straight line, either level or increasing at a constant rate or decreasing at a constant rate.

3. The velocity of the object changes at a constant rate (this third statement being obvious since the rate at which the velocity changes is the acceleration, which is assumed here to be constant).

4. The acceleration of the object at every instant is equal to the average acceleration of the object.

Describe a graph of velocity vs. clock time, assuming that the initial velocity occurs at clock time t = 0.

At what clock time is the final velocity then attained?

What are the coordinates of the point on the graph corresponding to the initial velocity (hint: the t coordinate is 0, as specified here; what is the v coordinate at this clock time? i.e., what is the velocity when t = 0?).

What are the coordinates of the point corresponding to the final velocity?

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Your solution:

The graph has a y axis labelled velocity. The x axis is clock time. There are two points on the graph. The intitial velocity coordinate is (0, 5m/s) The final velocity coordinate is (4, 25m/s)

confidence rating #$&*:

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Given Solution:

The initial velocity of 5 m/s occurs at t = 0 s so the corresponding graph point is (0 s, 5 m/s). The final velocity of 25 meters/second occurs after a time interval of `dt = 4 seconds; since the time interval began at t = 0 sec it ends at at t = 4 seconds and the corresponding graph point is ( 4 s, 25 m/s).

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: `q007. Is the v vs. t graph increasing, decreasing or level between the two points, and if increasing or decreasing is the increase or decrease at a constant, increasing or decreasing rate?

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Your solution:

The graph is increasing at a constant rate. The rise and run are positive so the slope is positive and therefore it is increasing. We are assuming uniform acceleration, so the increase is at a constant rate.

confidence rating #$&*:

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Given Solution:

Since the acceleration is uniform, the graph is a straight line. The graph therefore increases at a constant rate from the point (0, 5 m/s) to the point (4 s, 25 m/s).

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: `q008. What is the slope of the graph between the two given points, and what is the meaning of this slope?

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Your solution:

The slop is the rise/ run - in this case `dv/`dt = (25m/s - 5m/s)/4s = (20m/s)/4 s = 5m/s^2.

This slope is equal to average acceleration.

confidence rating #$&*:

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Given Solution:

The rise of the graph is from 5 m/s to 25 m/s and is therefore 20 meters/second, which represents the change in the velocity of the object. The run of the graph is from 0 seconds to 4 seconds, and is therefore 4 seconds, which represents the time interval during which the velocity changes. The slope of the graph is rise / run = ( 20 m/s ) / (4 s) = 5 m/s^2, which represents the change `dv in the velocity divided by the change `dt in the clock time and therefore represents the acceleration of the object.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: `q009. The graph forms a trapezoid, starting from the point (0,0), rising to the point (0,5 m/s), then sloping upward to (4 s, 25 m/s), then descending to the point (4 s, 0) and returning to the origin (0,0). This trapezoid has two altitudes, 5 m/s on the left and 25 m/s on the right, and a base which represents a width of 4 seconds. What is the average altitude of the trapezoid and what does it represent, and what is the area of the trapezoid and what does it represent?

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Your solution:

The average altitude is the sum of the y coordinates/2, in this case (5m/s +25m/s)/2 = (30m/s)/2 = 15 m/s This is also the average velocity.

To determine the area of the trapezoid we find the base time the average altitude. Thus

A= 4 s * 15m/s = 60 meters. This area is the distance travelled in the time interval.

confidence rating #$&*:

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Given Solution:

The two altitudes are 5 meters/second and 25 meters/second, and their average is 15 meters/second. This represents the average velocity of the object on the time interval. The area of the trapezoid is equal to the product of the average altitude and the base, which is 15 m/s * 4 s = 60 meters. This represents the product of the average velocity and the time interval, which is the displacement during the time interval.

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Self-critique (if necessary):OK - I should have used the term displacement instead of distance travelled.

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Self-critique rating:3

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Question: `q010. Students at this point often need more practice identifying which of the quantities v0, vf, vAve, `dv, a, `ds and `dt are known in a situation or problem. You should consider running through the optional supplemental exercise ph1_qa_identifying_quantities.htm . The detailed URL is http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_identifying_quantities.htm If you are able to quickly identify all the quantities correctly 'in your head', the exercise won't take long and it won't be necessary to type in any responses or submit anything. If you aren't sure of some of the answers, you can submit the document, answer and/or asking questions on only the problems of which you are unsure.

You should take a quick look at this document. Answer below by describing what you see and indicating whether or not you think you already understand how to identify the quantities. If you are not very sure you are able to do this reliably, indicate how you have noted this link for future reference. If you intend to submit all or part of the document, indicate this as well.

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Your solution:

The document is a set of 9 questions the later of which include some intriguing values like negative positions and cardinal directions. I understand how to identify quantities. I will not be submitting this optional assignment.

confidence rating #$&*:

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Given Solution:

You should have responded in such a way that the instructor understands that you are aware of this document, have taken appropriate steps to note its potential usefulness, and know where to find it if you need it.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: `q011. The velocity of a car changes uniformly from 5 m/s to 25 m/s during an interval that lasts 6 seconds. Show in detail how to reason out how far it travels.

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Your solution:

We can use the formula `ds = vAve/`dt.

We already know the `dt = 6 seconds so all we need to find is vAve.

vAve = (v0 + vf)/2 = (5m/s + 25 m/s)/2 = (30 m/s)/2 = 15m/s^2

Now we plug the values into our formula:

`ds = vAve/`dt = 15m/s^2 * 6s = 90 m

confidence rating #$&*:

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Question: `q012. The points (5 s, 10 m/s) and (10 s, 20 m/s) define a 'graph trapezoid' on a graph of velocity vs. clock time.

What is the average 'graph altitude' for this trapezoid?

Explain what the average 'graph altitude' means and why it has this meaning.

What is the area of this trapezoid? Explain thoroughly how you reason out this result, and be sure to include and explain your units.

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Your solution:

The average altitude is the sum of the y coordinates/2

(10 m/s + 20 m/s)/2 = (30 m/s)/2 = 15m/s

This average graph altitude is the y coordinate at the midpoint of the segment between our two points. It respresnts the average velocity.

The area of the trapezoid is the product of the base times the average altitude. The base in the time interval or change in clock time. This is 10s - 5 s = 5s

The area then is base times altitude which is 15m/s * 5s = 75 m

This area represents the displacement during the time interval.

confidence rating #$&*:

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Question: `q013. On a certain interval of duration `dt an object has initial velocity v_0 and final velocity v_f. In terms of the symbols v_0, v_f and `dt, what are the values of the following?

vAve

`dv

`ds

aAve

Be sure to explain your reasoning.

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Your solution:

vAve = (v_f + v_0)/2

This is so because to find the average velocity we take the sum of the inititial and final velocities and divide by 2.

`dv = v_f -v_0

The change in velocity is deteremined by subracting the initial velocity from the final velocity.

`ds = (v_f - v_0)/2 * `dt

The displacment is determined by taking the average velocity and multiplying it by the time interval.

aAve =(v_f -v_0)/`dt

The average acceleration is the rate of the change in velocity with respect to the change in clock time.

confidence rating #$&*:

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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. 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.

Questions related to q_a_

1. If we know the initial velocity v0, the final velocity vf and the time interval `dt for the motion of an object, then

How do we find its average acceleration for the interval?

How do we find its displacement during the interval?

How does each of your answers follow from the definition an average rate of change?

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

How do we construct a trapezoid to represent the motion on the intervening interval?

What aspect of the graph represents the change in velocity for the interval, and why?

What aspect of the graph represents the change in clock time for the interval, and why?

What aspect of the graph represents the acceleration for the interval, and why?

What aspect of the graph represents the displacement for the given interval, and why?

3. If we know v0, `dt and a for motion on an interval:

What quantity can we determine using the definition of acceleration?

What quantity can we then determine using the newly determined quantity with one of the given quantities?

At this point, what two quantities remain to be determined?

Which quantity can be determined using two of the five quantities we now know?

How can we use the six quantities we now know to find the value of the seventh?

Questions related to Introductory Problem Sets

1. In the figure below the given quantities v0, vf and `dt are listed in a straight line across the middle of the diagram.

There are lines from v0 and vf to vAve, and also from v0 and vf to `dv. Explain how v0 and vf are related to vAve, and also how they are related to `dv.

There are 'red' lines from vAve and `dt to `ds. How are vAve and `dt related to `ds by the definition of average velocity?

There are 'green' lines from `dv and `dt to a. How are `dv and `dt related to a by the definition of average acceleration?

ds_and_a_from_v0_vf_dt.gif (3439 bytes)

Questions related to Class Notes

1. If a cart is found to have acceleration 30 cm/s^2 when it rolls freely down an incline having slope .05, and acceleration 70 cm/s^2 on an incline having slope .08, then at what average rate is its acceleration changing with respect to ramp slope?

Questions/problems for Principles of Physics Students

4. About how long would it take an automobile moving at the speed limit to travel from Washington, D.C. to Cleveland, Ohio? You may base your answer on any reasonable estimate of the distance and the speed limit, but you should specify your estimates and show how you get from your estimates to your answer.

5. Estimate how long it would take a runner to run from New York to California. You may base your answer on any reasonable estimate of the distance and the running speed, but you should specify your estimates and show how you get from your estimates to your answer.

Questions/problems for General College Physics Students

6. Estimate how many liters of water a human being drinks in a lifetime. Would this water fill an average swimming pool?

Questions/problems for University Physics Students

What is the angle between the between the vectors -2i+6j and 2i - 3j ?

What is the cross product A X B of vector A, whose magnitude is 12.0 meters and which is directed along the negative y axis, and the vector B, with magnitude 18.0 m and directed at 37 deg as measured counterclockwise from the positive x axis? Give both the magnitude and the direction of the cross product.