qa04

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course PHY 241

It’s average velocity so it would be aAve.

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Average velocity is vAve.

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INSTRUCTOR RESPONSE:

Good, but note:

It’s average acceleration (not average velocity) so it would be aAve.

In most of your course acceleration is constant, so initial accel = final accel = aAve.

• In this case we can just use 'a' for the acceleration.

STUDENT QUESTION

If I understand this correctly, the average rate in which velocity changes is acceleration???? Where did average

acceleration fit into the problem, the problem asked for the average rate that velocity changed?

INSTRUCTOR RESPONSE

Acceleration is rate of change of velocity with respect to clock time. So the terms 'average acceleration' and 'average rate of change of velocity with respect to clock time' are identical. The term 'average rate of change of velocity' actually leaves off the 'with respect to clock time', but in the context of uniformly accelerated motion 'average rate of change of velocity' is understood to mean 'average rate of change of velocity with respect to clock time' .

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Self-critique (if necessary):

ok

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Self-critique rating:3

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Question: `q006. If a runner is moving at 6 meters / sec at clock time t = 1.5 sec after starting a race, and at 9 meters / sec at clock time t = 3.5 sec after starting, then what is the average acceleration of the runner between these two clock times?

If you can, answer the question as posed. If not, first consider the two questions below:

• What is the change `dv in the velocity of the runner during the time interval, and what is the change `dt in clock time during this interval?

• What therefore is the average rate at which the velocity is changing with respect to clock time during this time interval?

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Your solution:

6m/s at 1.5 s

9m/s at 3.5s

‘dv = 9m/s-6m/s = 3m/s

‘dt= 3.5s-1.5 s = 2s

aAve=’dv/’dt=3m/s/2s = 1.5m/s/s

confidence rating #$&*:

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Given Solution:

We see that the runner's velocity changes from 6 meters/second to 9 meters/second, a change of `dv = 9 m/s - 6 m/s = 3 m/s, during a time interval lasting from t = 1.5 sec to t = 3.5 sec.

The duration of the interval is `dt = 3.5 s - 1.5 s = 2.0 s.

The rate at which the velocity changes is therefore 3 m/s / (2.0 s) = 1.5 m/s^2.

STUDENT QUESTION

I'm not understanding why you have the power of 2 for.

INSTRUCTOR RESPONSE

When you divide m/s by s you do the algebra of the fractions and get m/s^2. You don't get m/s.

The distinction is essential:

• m/s^2 is a unit of acceleration.

• m/s is a unit of velocity.

Velocity and acceleration are two completely different aspects of motion.

The algebra of dividing m/s by s was given in a previous question in this document. In a nutshell, (m/s) / s = (m/s) * (1/s) = m/s^2.

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Self-critique (if necessary):

ok

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Self-critique rating:3

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Question: `q007. On a graph of velocity vs. clock time, we can represent the two events of this problem by two points on the graph. The first point will be (1.5 sec, 6 meters/second) and the second point will be (3.5 sec, 9 meters / sec). What is the run between these points and what does it represent?

What is the rise between these points what does it represent?

What does the slope between these points what does it represent?

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Your solution:

The run is the time interval of 2seconds.

The rise is the change in velocity of 3 m/s

The slope is the acceleration or rise over run of 3/2 =1.5 m/s^2

confidence rating #$&*:

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Given Solution:

The rise from the first point to the second is from 6 meters/second to 9 meters/second, or 3 m/s. This represents the change `dv in velocity. The run is from 1.5 seconds to 3.5 seconds, or 2 seconds, and represents the change `dt in clock time. The slope, being the rise divided by the run, is 3 m/s / (2 sec) = 1.5 m/s^2. This slope represents `dv / `dt, which is the average acceleration during the time interval. You may if you wish comment on your understanding to this point.

STUDENT QUESTION

Are we going to use the terms acceleration and average acceleration interchangeably in this course? I just want to make sure

I understand.

INSTRUCTOR RESPONSE

Good question.

The term 'acceleration' refers to instantaneous acceleration, the acceleration at a given instant.

The term 'average acceleration' refers to the average acceleration during an interval, calculated by subtracting initial from final velocity and dividing by the change in clock time.

If acceleration is uniform, it's always the same. If acceleration is uniform, then, it is unchanging. In that case the instantaneous acceleration at any instant is equal to the average acceleration over any interval.

So when acceleration is uniform, 'acceleration' and 'average acceleration' are the same and can be used interchangeably.

Acceleration isn't always uniform, so before using the terms interchangeably you should be sure you are in a situation where acceleration is expected to be uniform.

This can be visualized in terms of graphs:

The instantaneous acceleration can be represented by the slope of the line tangent to the graph of v vs. t, at the point corresponding to the specified instant.

The average acceleration can be represented by the average slope between two points on a graph of v vs. t.

If acceleration is uniform then the slope of the v vs. t graph is constant--i.e., the v vs. t graph is a straight line, and between any two points of the straight line the slope is the same. In this case the tangent line at a point on the graph is just the straight-line graph itself.

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Self-critique (if necessary):

ok

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Question: `q008. In what sense does the slope of any graph of velocity vs. clock time represent the acceleration of the object? For example, why does a greater slope imply greater acceleration?

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Your solution:

Since slope is calculated using the same information that we would use to calculate acceleration the two will be identical.

confidence rating #$&*:

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Given Solution:

Since the rise between two points on a graph of velocity vs. clock time represents the change in `dv velocity, and since the run represents the change `dt clock time, the slope represents rise / run, or change in velocity /change in clock time, or `dv / `dt. This is the definition of average acceleration.

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Self-critique (if necessary):

ok

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Self-critique rating:2

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Question: `q009. This is the same situation as in the preceding problem: An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of velocity vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).

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Your solution:

Increasing at a constant rate up to a certain velocity and then increasing at a decreasing rate.

confidence rating #$&*:

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Given Solution:

Your graph should have velocity as the vertical axis and clock time as the horizontal axis. The graph should be increasing since the velocity starts at zero and increases. At first the graph should be increasing at a constant rate, because the velocity is increasing at a constant rate. The graph should continue increasing by after a time it should begin increasing at a decreasing rate, since the rate at which the velocity changes begins decreasing due to air resistance. However the graph should never decrease, although as air resistance gets greater and greater the graph might come closer and closer to leveling off. Critique your solution by describing or insights you had or insights you had and by explaining how you now know how to avoid those errors.

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Self-critique (if necessary):

Ok This was a straightforward question if you can visualize what the car is doing I had no difficutly with this problem.

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Self-critique rating:3

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Question: `q010. An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of acceleration vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).

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Your solution:

At first the graph would be constant since acceleration is constant, until the certain point in which it will begin decreasing at an increasing rate.

confidence rating #$&*:

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Given Solution:

Your graph should have acceleration as the vertical axis and clock time as the horizontal axis. At first the graph should be neither increasing nor decreasing, since it first the acceleration is constant. Then after a time the graph should begin decreasing, which indicates the decreasing rate at which velocity changes as air resistance begins having an effect. An accurate description of whether the graph decreases at a constant, increasing or decreasing rate is not required at this point, because the reasoning is somewhat complex and requires knowledge you are not expected to possess at this point. However it is noted that the graph will at first decrease more and more rapidly, and then less and less rapidly as it approaches the t axis.

STUDENT QUESTION: Can you clarify some more the differences in acceleration and velocity?

INSTRUCTOR RESPONSE: ** Velocity is the rate at which position changes and the standard units are cm/sec or m/sec.

Acceleration is the rate at which velocity changes and its standard units are cm/s^2 or m/s^2.

Velocity is the slope of a position vs. clock time graph.

Acceleration is the slope of a velocity vs. clock time graph. **

STUDENT QUESTION:

In the problem it states that velocity continues to increase even though the rate at which velocity changes decreases.

I don’t understand your the slope will decrease if this is true. I can understand a diminish in velocity and time, but not a down turn of the slope, which is what your solution leans to.

INSTRUCTOR RESPONSE

Your thinking is good, but you need carefully identify what it is you're describing.

The question here concerns the acceleration vs. clock time graph, whereas most of your comments apply to the velocity vs. clock time graph.

Under these conditions the slope of the velocity vs. clock time graph will decrease, this will occur as long as the acceleration vs. clock time graph decreases, regardless of whether that decrease is at a constant, an increasing or a decreasing rate.

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Self-critique (if necessary):

ok

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Self-critique rating:3

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: `q011. Which changes its velocity more quickly, on the average, a car which speeds up from 50 mph to 60 mph in 5 seconds, or a car which speeds up from 5 mph to 25 mph in 6 seconds?

Be sure to explain your answer.

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Your solution:

The car that accelerated from 50 to 60 mph had a ‘dv= 60mph-50mph=10mph ‘dt= 5s

aAve=`dv/dt=10mph/5s=2mph/s

The car that accelerated from 5 to 25 mph had a ‘dv= 25mph-5mph=20mph ‘dt= 6s

aAve=`dv/dt=20mph/6s=3.3 mph/s

The second car changes velocity more quickly.

confidence rating #$&*:

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Question: `q012. What do we get if we divide 40 meters by 5 meters / second?

Your answer will include a number and its units. You should explain how you got the units of your answer.

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Your solution:

40m / 5m/s flip the fraction and multiply

40m * s/5m meters cancel out

8s

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Question: `q013. On a graph of velocity v vs. clock time t we have the two points (5 s, 10 m/s) and (10 s, 20 m/s). What is the average slope of the graph between these points, and what does this average slope mean?

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Your solution:

Average slope is the average acceleration of an object following this graph.

Slope = `dy/`dx= (y2-y1)/(x2-x1)=(20m/s-10m/s)/(10s-5s)

10m/s/5s

2m/s/s

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Right result but you haven't explained the meaning of the average slope.

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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. 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_

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&#Good responses. See my notes and let me know if you have questions. &#