cq_1_031

Your 'cq_1_03.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

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30 cm/s is not a velocity. The only thing you know at this point about velocity is that the average velocity is 6 cm/s.

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If you can learn the subject without using these questions, then you're welcome to do so. Most students need them. However you're doing better at this point than most students, and if I was to select a handful of students who might be able to do without them you would be among them.

However I would be hesitant to actually recommend this for anyone who makes errors on the 'seed' questions, or who requires more than a few minutes to solve them (you've made some errors and have taken more than a few minutes on some of the questions, though on most you have worked quickly and accurately).

The original structure of this question mislead you into an extraneous step, in which you concluded that the average velocity was 15 cm/s. Your original answer to this question was 6 cm/s and this is the correct average velocity on which the rest of the problem is based. This would have led you to the correct solution:

ave vel = 6 cm/s

init vel is zero and accel is uniform so vf = 12 cm/s

change in vel is `dv = 12 cm/s - 0 cm/s = 12 cm/s

ave accel is `dv / `dt = 12 cm/s / (5 s) = 2.4 cm/s^2.

See the following more detailed discussion, which includes a lot of important connections as well as some general advice and a list of common pitfalls.

The following is a solution to the given problem. Please see the given solution and submit a self-critique of your solution, by copying your solution into a text editor and inserting revisions, questions, and/or self-critiques, marking your insertions with ####. 

Submit using the Submit Work Form.

Among other things this problem should illuminate the difference between

midpoint clock time, change in clock time, clock time, time interval, initial

velocity, final velocity, average velocity, change in velocity and the meaning

of slope. 

A ball starts with

velocity 0 and accelerates uniformly down a ramp of length 30 cm, covering the distance in

5 seconds.

• What is its average

  velocity? 

Its average velocity is the average rate at

which position changes with respect to clock time, which in turn is equal to

(change in position) / (change in clock time), so we have

• vAVe = (30 cm) / (5 sec) = 6 cm / sec.

• If the acceleration of the

  ball is uniform then its average velocity is equal to the average of its

  initial and final velocities. 

You know its average

velocity, and you know the initial velocity is zero. 

What therefore must be

the final velocity? 

The initial velocity of the ball is 0 and

acceleration is uniform.  Therefore its final velocity is

• vf = 2 * vAVe = 2 * (6 cm / sec) = 12

  cm/s.

• By how much did its

  velocity therefore change? 

The velocity of the ball changed from its

initial velocity v0 = 0 cm/s to its final velocity vf = 12 cm/s.  So

the change in velocity is

• `dv = vf - v0 = 12 cm/s - 0 cm/s = 12

  cm/s.

• At what average rate did

  its velocity change with respect to clock time? 

The average rate of change of velocity with

respect to clock time is by definition

• ave rate =

(change in velocity) / (change in clock

time) =

12 cm/s / (5 s) =

2.4 cm/s^2.

• What would a graph of its

  velocity vs. clock time look like?  Give the best description you can.

The reasoning here is as follows:

• Acceleration is uniform, meaning it is

  unchanging. 

• Acceleration is the rate of change of

  velocity with respect to clock time, which is represented by the slope

  of the v vs. t graph. 

• The slope of the graph is therefore

  constant. 

• The graph of v vs. t rises along a

  straight line from v = 0 to v = 30 cm/s while clock time t changes by 5

  seconds. 

• The rise of the graph between these

  points is 30 cm/s, the run is 5 s, and the slope of the line is rise /

  run = (30 cm/s) / (5 s) = 6 cm/s^2. 

• The slope is identical to the (uniform)

  acceleration. 

Additional Suggestions

Any uniformly accelerated motion problem can be reasoned out from the

three things listed below.

On every problem you should start by writing down all three, and you should

continue writing down these definitions, and applying them very carefully,

until you are very sure of them. 

• the definition of average

  velocity

• the definition of average acceleration.

• a short note that if acceleration is uniform, then average velocity

  is the average of the initial and final velocities

To answer the question 'by how much does velocity change' on an interval

you should ask yourself the following three questions:

• What was the velocity at the beginning of the interval?

• What was the velocity at the end of the interval?

• What therefore was the change in velocity?

To apply the definition of the average rate of change of velocity with

respect to clock time:

• What is this definition?

• What is the change in velocity?

• What is the change in clock time?
   

It is important to distinguish the graph of velocity vs. clock time from

the graph of position vs. clock time. 

• If acceleration is uniform, the graph of v vs. t is a straight line. The slope

  of this line is the acceleration.

• The graph of position vs. t has slopes that represent velocities. If

  acceleration is constant (and not zero) velocities are continually

  changing, and the slopes of this graph change accordingly, giving the

  graph a curved shape.

Don't confuse average velocity with change in velocity or with average

rate of change of velocity with respect to clock time.

• Average velocity is change in position / change in clock time. 

• If the v vs. t graph on an interval is a straight line segment, the

  average velocity on that interval occurs at the midpoint of the segment.

• If the v vs. t graph on an interval is a straight line, then the

  average velocity is equal to the average of the initial and final

  velocities,

vAve = (final velocity + initial velocity) / 2.

• The change in velocity on an interval is

`dv = (final velocity) - (initial velocity). 

This quantity is unrelated to the average velocity. 

• The average rate of change of velocity with respect to clock time on

  an interval follows from the definition of rate of change, and is given

  by

ave rate = (final velocity - initial velocity) / (change in

clock time)

This quantity is be definition equal to the average acceleration.

• In summary, don't confuse the three quantities average velocity,

  change in velocity and average acceleration.  Know which is which

  and know which applies to the question you are answering or the

  situation you are analyzing:

• vAve = `ds / `dt, which also is equal to (vf + v0) / 2 if

  acceleration is uniform

• `dv = vf - v0

• aAve = (vf - v0) / `dt