query 5

course phy201

9:09 pm 2/13/2010

Query 5 005. 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 9 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 the subject of the next few questions as well)

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Your solution:

25m/s - 5m/s = 20m/s is how much it changes

aAve= 'dv * 'dt

aAve= 20m/s * 4s = 80m/s^2

vave= (25m/s + 5 m/s)/2 = 15 m/s

confidence rating #$&*:2

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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):

ooops I had my formula wrong for aAve should be aAve = 'dv/'dt

aAve= 20m/s/4s = 5m/s^2

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Self-critique rating #$&*:3

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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

'ds = 15 m/s * 4s = 60m

confidence rating #$&*:2

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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, my confidence is alittle low on these problems since I was getting the vave and the change in velocity mixed

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Self-critique rating #$&*:3

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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:

the acceleration is (change in velocity)/(change in clocktime)

the distance = the average velocity * clocktime

So to get the acceleration we must first determine the change in velocity and divide that by change in time. To get the distance we must average the velocities and mulitply by the clocktime

confidence rating #$&*: 2

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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):

I uderstand this

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Self-critique rating #$&*:3

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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:

acceleration = (vf- V0)/'dt

Vave= (vf+V0)/2

'ds= (Vf+v0)/2 * 'dt

confidence rating #$&*:3

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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.

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Self-critique (if necessary):

I got it

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Self-critique rating #$&*:3

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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.

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.

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 o velocity vs clocktime would have the velocity on the vertical axis and the clocktime on the horizontal axis. This will form a straight line.

The final velocity is attained at 4 seconds

when t = 0 intitial velocity is 5 cm/s

4s, 25cm/s is the coordinates of the points corresponding to the final velocity

confidence rating #$&*:2

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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, I understand the coordinates of this graph

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Self-critique rating #$&*:3

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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 v vs t graph is increasing between the two points the slope of this graph is 20m/s / 4s = 5m/s^2 same as the aAve so since the graph is increasing at an increasing rate.

confidence rating #$&*:2

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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):

So if the acceleration is uniform and the graph is a stright line would that be level between the two points, with a rate that is slowly increasing?????????? Would the graph have no steepness?????????????

A straight line can slope upward or downward. It can be very steep, or only a little bit steep, or it can be level (with no steepness); it just can't change its slope.

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Self-critique rating #$&*:2

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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 slope is 20m/s / 4s = 5 m/s^2

the slope is the same as the acceleration rate for this graph

confidence rating #$&*:2

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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, that is what I thought it was, could this alos be the the average rate of change??????

Right, but you have to specify which average rate of change.

This would be the average rate of change of velocity with respect to clock time.

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Self-critique rating #$&*:3

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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 25m/s + 5 m/s /2 = 15m/s

this is the same as the average rate of velocity

confidence rating #$&*:3

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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, so the area is = to vave * 'dt which is the same as the displacement

I did not realize before that the area and 'ds were the same

They are the same because of the meanings of the average 'graph altitude' and the 'graph width', and the fact that you multiply to two to get the area.

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Self-critique rating #$&*:3

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