Asst_4_QA4

#$&*

course PHY 201

2/06 8

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

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

aAve= 'dv / 'dt = (25 - 5 m/s) / 4sec = 20 m /s / 4 sec = 5 m/s/sec

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A vehicle will change its velocity at a rate relative to the amount of pressure exerted on the

accelerater pedal, or, how much fuel is allowed into the engine by the throttle, which effects the

rate at which the engine combusts and forces the pistons up and down creating the force to

drive the transmission at greater speeds or rates.

The more force exerted on the pedal, the more fuel is fed to the engine, the more combustion,

the faster speeds obtained by the vehicle.

If a supercar weighing less than my Jeep, but with twice the engine power starts next to mine

and we both slam our pedals to the floor, the supercar will accelerate at a faster rate because

its engine has more capabilty to convert fuel to power than mine.

Likewise a V8 Jeep next to my V6 would over a longer clocktime reach greater rates of

velocity change because it has a larger engine. In the short term, timing and auto mechanics

even the field, however on simple level terms a more powerful engine is capable to greater

rate of velocity change.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Velocity is the change is displacement, in this case measured in meters, divided by the change

in clock time. This gives us a measurment divided by a time increment, m/sec

Rate of Change of Velocity calculates the change in velocities of two points of clock time, or

aAve ='dv/'dt. Since velocity is measured in terms of m/s and clock time is secs, 'dv / 'dt would

give us m/s/s.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(meters / sec) * (1 / sec) = (meters * 1) / (sec * sec) or simply, meters/sec^2

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

aAve = 'dv /'dt = (-5 - 10 m/s) / 5sec = -15 m/s/5s = -3m/s/s or -3m/s^2

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

aAve = 'dv/ 'dt

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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' .

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

aAve= (9-6 m/s)/(3.5 - 1.5 sec)= 3m/s/ 2sec = 3/2 m/s^2 or 1.5m/s^2

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The run represents the change in clocktime from 1.5 sec to 3.5 sec, or just 2 seconds.

The rise is the change in velocity from 6m/s to 9m/s, or just 3m/s.

The slope between these two intervals represents the acceleration (average) for these events.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Acceleration is defined as the rate of change between intervals of velocity and clocktime. If

the first interval was (3sec , 3m/s) and the second was (5sec, 7m/s), my aAve would be (7m/s

-3m/s) / (5sec - 3sec) = 4m/s / 2sec = 2m/s/s.

Greater slope is yielded when clocktime of an interval on a velocity v. clocktime graph is

shortened or when velocity is increased. A steeper graph line represents a faster aAve.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

My graph is increasing at a constant rate for an unkown interval and then changes to an

increasing at a decreasing rate.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

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).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

My graph shows a line on a velocity v clock time graph that is constant and then gradually

decreases.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.