QA Assignment 4

#$&*

course PHY 121

6/13 7

004. Acceleration

Goals:

Definition of average rate of change of velocity with respect to clock time.

Apply the definition of average rate of change to define acceleration.

Interpret the slope of a velocity vs. clock time graph

(understand the defining characteristic of the v vs. t graph for constant acceleration:

not well emphasized at this point)

Describe the v vs t graph of an object with nonuniform acceleration

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

At a certain instant the speedometer of a car reads 5 meters / second (of course cars

in this country generally read speeds in miles per hour and km per hour, not meters /

sec; but they could easily have their faces re-painted to read in meters/second, and we

assume that this speedometer has been similarly altered). Exactly 4 seconds later the

speedometer reads 25 meters/second (that, incidentally, indicates very good

acceleration, as you will understand later). At what average rate is the speed of the

car changing with respect to clock time?

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Your solution:

The change in velocity is 25 m/s - 5 m/s = 20 m/s.

To figure the rate that the speed is changing, you take the 20 m/s and divide by the

elapsed time of 4 seconds. This give you a change of 5 meters per second per second.

confidence rating #$&*:

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Given Solution:

The rate of change of the speed with respect clock time is equal to the change in the

speed divided by the change in the clock time. So we must ask, what is the change in

the speed, what is the change in the clock time and what therefore is the rate at which

the speed is changing with respect to clock time?

The change in speed from 5 meters/second to 25 meters/second is 20 meters/second. This

occurs in a time interval lasting 4 seconds. The average rate of change of the speed is

therefore (20 meters/second)/(4 seconds) = 5 meters / second / second. This means that

on the average, per second, the speed changes by 5 meters/second.

STUDENT QUESTION

Would we not have s^2 since we are multiplying s by s?

INSTRUCTOR RESPONSE

That is correct. However in this question I've chosen not to confuse the issue by

simplifying the complex fraction m/s/s, which we address separately.

To clarify, m / s / s means, by the order of operations, (m/s) / s, which is (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: `q002. Explain in commonsense terms of the significance for an automobile of

the rate at which its velocity changes. Do you think that a car with a more powerful

engine would be capable of a greater rate of velocity change?

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Your solution:

Accelerating too fast can put a lot of stress on the motor. It's better for the car to

accelerate at a more gradual rate. Because of its more powerful engine, the car with a

more powerful engine is able to change its velocity at a greater rate.

confidence rating #$&*:

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Given Solution:

A car whose velocity changes more rapidly will attain a given speed in a shorter time,

and will be able to 'pull away from' a car which is capable of only a lesser rate of

change in velocity. A more powerful engine, all other factors (e.g., weight and

gearing) being equal, would be capable of a greater change in velocity in a given time

interval.

STUDENT COMMENT:

The significance for an automobile of the rate at which its velocity changes is the

amount of speed it takes to travel to a place in a certain amount of time. If one car is

traveling along side another, and they are going to the same location the velocity will

be how long it take this car to get to this location going at a speed other than the

other car. If a car with a more powerful engine were to travel the same distance its

velocity would be capable of a greater rate if increased speed occurred.

INSTRUCTOR RESPONSE:

It's necessary here to distinguish between velocity, which is a pretty intuitive

concept, and rate of change of velocity, which is much less intuitive and less familiar.

An object can change velocity at a constant rate, from rest to a very high velocity. All

the while the rate of change of velocity with respect to clock time can be unchanging.

So the rate of change of velocity with respect to clock time has nothing to do with how

fast the object is moving, but rather with how quickly the velocity is changing.

Moving from one location to another, the displacement is the change in position. If the

displacement is divided by the time required we get the average rate of change of

position with respect to clock time, or average velocity.

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Self-critique (if necessary): OK

I understand that accelerating more quickly will allow a car to ""pull away"" from a stop

or another car more quickly than a car that it not accelerating quickly.

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Self-critique rating: 3

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Question: `q003. Explain how we obtain the units meters / second / second in our

calculation of the rate of change of the car's speed.

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Your solution:

You find the change in speed between two data points. In order to do this, you need to

subtract the first speed from the 2nd speed. So, if the change in speed was from 10 m/s

to 20 m/s, you are working with a difference of 10 m/s. Then you divide this by the

number of seconds it took to make that change in speed. So if it took 5 seconds to go

form 10 m/s to 20 m/s, you would divide the 10 m/s difference by 5 to find out that the

change in speed is 10/5 or 2 meters per second every second.

confidence rating #$&*:

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Given Solution:

When we divide the change in velocity, expressed in meters/second, by the duration of

the time interval in seconds, we get units of (meters / second) / second, often written

meters / second / second.

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Self-critique (if necessary): OK

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Self-critique rating: 3

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Question: `q004. The unit (meters / second) / second is actually a complex fraction,

having a numerator which is itself a fraction. Such a fraction can be simplified by

multiplying the numerator by the reciprocal of the denominator. We thus get (meters /

second) * (1/ second). What do we get when we multiply these two fractions?

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Your solution:

To multiply fractions, you simply multipy the 2 numerators, which would give us 1 * m,

or the meters. Then you multiply the 2 denominators, which would be seconds * seconds,

which would be s^2. This would give you the number of meters/seconds^2.

confidence rating #$&*:

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Given Solution:

Multiplying the numerators we get meters * 1; multiplying the denominators we get second

* second, which is second^2. Our result is therefore meters * 1 / second^2, or just

meters / second^2. If appropriate you may at this point comment on your understanding

of the units of the rate of change of velocity.

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Self-critique (if necessary): OK

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Self-critique rating: 3

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Question: `q004. If the velocity of an object changes from 10 m/s to -5 m/s during a

time interval of 5 seconds, then at what average rate is the velocity changing with

respect to clock time?

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Your solution:

To change from 10 m/s to -5 m/s, it is a change of -15 m/s. If the change is made over

a time interval of 5 seconds, the average rate of change is -15/5, which is -3 m/s/s.

confidence rating #$&*:

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Given Solution:

We see that the velocity changes from 10 meters/second to -5 meters/second, a change of

-15 meters / second, during a five-second time interval. A change of -15 m/s during a 5

s time interval implies an average rate of -15 m/s / (5 s) = -3 (m/s)/ s = -3 m/s^2.

This is the same as (-3 m/s) / s, as we saw above. So the velocity is changing by -3

m/s every second.

STUDENT QUESTION

Do you have to do the step -3 m/s /s. Because I get the same answer not doing that.

INSTRUCTOR RESPONSE

Your solution read ' -5 m/s - 10 m/s = -15 m/s / 5 seconds = -3 m/s '.

Everything was right except the units on your answer. So the answer to you question is

'Yes. It is very important to do that step.'

The final answer in the given solution is '-3 m/s every second', which is not at all the

same as saying just '-3 m/s'.

-15 m/s / (5 s) = -3 m/s^2, which means -3 m/s per s or -3 cm/s every second or -3

m/s/s.

-3m/s is a velocity. The question didn't ask for a velocity, but for an average rate of

change of velocity.

-3 m/s per second, or -3 m/s every second, or -3 m/s/s, or -3 m/s^2 (all the same) is a

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: `q005. You should have noted that velocity can be positive or negative, as

can the change in velocity or the rate at which velocity changes. The average rate at

which a quantity changes with respect to time over a given time interval is equal to the

change in the quantity divided by the duration of the time interval. In this case we

are calculating the average rate at which the velocity changes. If v represents

velocity then we we use `dv to represent the change in velocity and `dt to represent the

duration of the time interval. What expression do we therefore use to express the

average rate at which the velocity changes?

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Your solution:

'dv/'dt

You divide the change in velocity by the amount of time it takes to make that change.

confidence rating #$&*:

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Given Solution:

The average rate would be expressed by [ave rate of velocity change with respect to

clock time] = `dv / `dt. The expression [ave rate of velocity change with respect to

clock time] is pretty cumbersome so we give it a name. The name we give it is 'average

acceleration', abbreviated by the letter aAve. Using a to represent acceleration, write

down the definition of average acceleration. The definition of average acceleration is

aAve = `dv / `dt. Please make any comments you feel appropriate about your

understanding of the process so far.

STUDENT COMMENT:

It’s average velocity so it would be aAve.

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:

1.5 m/s/s

(9m/s-6m/s)/(3.5 s - 1.5 s) = (3 m/s)/2 seconds

3 Divided by 2 is 1.5

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 9-6=3. It is the difference in the meters per second at those points in

time.

The rise is 3.5-1.5=2. It is the difference in the time it took to get from 6 m/s to 9

m/s.

The slope of 3/2 or 1.5 is that for every 1.5 meters run, it took 1 second.

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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Self-critique rating: 3

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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:

The slope is how much distance is covered in a given amount of time. When more distance

is covered in a shorter amount of time, the slope of the line will be greater. It shows

the acceleration of the object because it indicates how far it has gone and how long it

took.

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: 3

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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:

The first part of the graph has the slope increasing at a constant rate. At the the

point where the air resistance has an effect on the velocity, the slope of the line

decreases, although the line continues to increase at a rate that is constant.

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):

I did not describe the x axis as time and the y axis as distance, althought it is how I

set up my graph. I can see that the way I drew the second part of my line, it does

increase at a decreasing rate. I did not get to the point where it leveled off. I did

not start my graph at the velocity of zero because it did not say that it was at zero

when it started.

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Self-critique rating: 2

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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:

The first part of the graph has the slope increasing at a constant rate. At the the

point where the air resistance has an effect on the velocity, the slope of the line

continues to increase but at a decreasing rate. The x axis shows time and the y axis

shows acceleration.

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):

I was still thinking in terms of velocity, so I was confused about the beginning of the

graph being horizontal and then decreasing. I see now that if the car is accelerating

at a constant speed and the y axis is acceleration, then the line will be horizontal to

the x axis at the point that shows the constant acceleration. For example, if the car

is accelerating at the rate of 3 miles per hour for the first 10 minutes, then the line

on the graph will go across the 3 mph for 10 min. Then, if it drops to 2 mph for 3

minutes, the graph will drop and level off for that until the point where the car is no

longer accelerating.

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Self-critique rating: 3

If you understand the assignment and were able to solve the previously given problems

from your worksheets, you should be able to complete most of the following problems

quickly and easily. If you experience difficulty with some of these problems, you will

be given notes and we will work to resolve difficulties.

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Question: `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 change in the velocity of the first car is 60-50 = 10 mph. The time change is 5

seconds. The first car increases its speed at the rate of 10 mph/ 5 sec, which is 2

miles per second.

The change in the velocity of the second car is 25-5 = 20 mph. The time change is 6

seconds. The second car increases its speed at the rate of 20 mph in 6 seconds, which

is 3.3 mph/sec.

So the second car is changing its velocity more quickly than the first car.

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:

If you divide 40 meters by 5 m/s, you will get 8 seconds.

40 meters divided by 5 m/sec becomes 40/1 meters times 1 sec/5 m. Then you multiply the

fractions. The meters cancel each other out when you do: (40 m * 1 sec)/(1 * 5 m).

This gives you 40/5 seconds, or 8 seconds.

confidence rating #$&*:

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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:

The average slope of the line is 2. This means that it took 1 second for the object to gain 2 meters per second increase in velocity. Or a rise of 2 means for every 2 m/s increase, the run 1 was 1 or 1 second elapsed time.

@&

Good. This means that velocity changes at a rate of 2 m/s per second, or 2 m/s/s or 2 m/s^2. This is the acceleration.

*@

The rise is obtained by subtracting the y coordinates 20-10 to get 10. The run is obtrained by subtracting the x coordinates 10-5 to get 5. The slope is rise / run or 10/5, which is 2.

confidence rating #$&*:

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Self-critique (if necessary):

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Self-critique rating:

&#Your work looks good. See my notes. Let me know if you have any questions. &#