phy201
Your calculation is correct, as are your units, so you did very well on the
question.Instead of
r = d / t,
you should be thinking
ave rate = change in A / change in B.
The change in A is sometimes a distance, in which d could be an
appropriate letter, but more often it is something else, as in this example.'Distance' is not the most commonly used quantity ussed in analyzing
motion; rather displacement is the more appropriate quantity.Since the letter 'd' is more likely to invoke the word 'distance', it is
likely to be misleading.
So for example the expression `dy / `dx stands for 'change in y
/ change in x', which by definition of average rate of change stands for the average rate of change of y with respect to x.
At the level of Principles of Physics and even more so in General College Physics we consider the concept of the instantaneous rate of change, which occurs when `dx gets smaller and smaller, approaching zero. The limiting value of `dy / `dx, and `dx shrinks to zero, is universally represented by the expression dy / dx.In University Physics we take this a step further with the concept of the
derivative function dy/dx.
The symbol t generally stands for the 'running time' on a timekeeping
device (e.g., a clock), and not for the interval between to events that occur at different clock times.To use the same symbol t for the duration of the time interval during
which the distance is covered is inherently confusing.
The formula r = d / t should therefore be replaced in our minds by a more
general formula likewhere x is position and t is clock time.
`dx is literally read as 'delta x' or 'change in x'.
- If x represents position, then `dx represents change in position.
- `dt is read as 'change in t', so if t represents clock time, `dt represents change in clock time.
Thus r in this context would stand for 'change in position / change in
clock time', which by definition of
average rate is the average rate of change of position with respect to clock time.
The current question does not deal with change in position, but with change
in velocity. In this case it would be moreWe said above that the use of r for rate is 'pretty much' OK. However we are
already running into a problem with the use ofWe previously used r for the average rate of change of position with
respect to clock time, r = `dx / `dt.
Now we are using r for the average rate of change of velocity with respect to clock time, r = `dv / `dt.For the present we'll just make note of this ambiguity. Shortly we'll
resolve it.
Now let's return to your solution.
Your calculation for this problem, which I repeat arrived at the correct
final result using correct reasoning and including correct units (when 80% of students in your course at this stage at least fail to do the units calculation correctly, and most do not spell out their reasoning) was presented as follows:r=d/t
d=10cm/s - 40cm/s = - 30cm/s
t=3s
r=30cm/s / 3s
r=10cm/s^2.
This is a very good solution but would be clearer if we used different
symbols. The same solution, using `dv instead of dr =`dv / `dt
`dv = 10cm/s - 40cm/s = -30cm/s
`dt = 3s
r = - 30cm/s / 3s
r = - 10cm/s^2.
It should be clear why the notation used in this revision is more specific
and clearer than the notation of the r = d / t formula.The quantity we calculate here is `dv / `dt, the average rate of change
of velocity with respect to clock time.
We could use the abbreviation 'ave roc of v wrt t' instead of just r. With this notation the solution would readave roc of v wrt t =`dv / `dt
`dv = 10cm/s - 40cm/s = -30cm/s
`dt = 3s
ave roc of v wrt t = - 30cm/s / 3s
ave roc of v wrt t = - 10cm/s^2.This clearly defines the nature of the average rate we are calculating.
With this notation we have a complete and specific solution to the problem.
Since we've come this far, we have a good opportunity to go a little further
and use a name for the average rate of change of velocity with respect to clock time.We will call this quantity the 'average acceleration'. Average
acceleration is defined to be the average rate of change of velocity with respect to clock time.
Using the notation a_Ave for this quantity your solution becomesa_ave =`dv / `dt
`dv = 10cm/s - 40cm/s = -30cm/s
`dt = 3s
a_Ave = - 30cm/s / 3s
a_Ave = - 10cm/s^2.
Since we've gone this deeply into the discussion, one other idea worth noting
is that in most of the situations we encounter in the early part of a first-semester physics course, we expect acceleration to be uniform, unchanging, for the duration of some extended time interval.If this is the case, then the average acceleration is the same as the
initial acceleration or the final acceleration, or the acceleration at any other instant during that interval. In this case we can simply drop the subscript 'Ave' and use the letter a for 'the constant acceleration'.
In your problem it was not specified that the acceleration is constant. So the best you can say from the given information is that a_Ave = - 10 cm/s^2.
Had it been specified that acceleration is constant, then your solution could have readacceleration is constant, so a_Ave can be represented simply by the
letter a
a_ave = `dv / `dt, so
a = `dv / `dt.
`dv = 10cm/s - 40cm/s = -30cm/s
`dt = 3s
a = - 30cm/s / 3s
a = - 10cm/s^2.
One final thing it worth noting:
The average slope of the v vs. t graph, between two graph point,
represents an average acceleration. If acceleration is constant, the v vs. t graph, which represents velocity vs. clock time, has constant slope so the graph is a straight line. The converse is also true: If the v vs. t graph is a straight line, then acceleration is constant.
All these ideas will be developed in this assignment and in the next few
assignments. This discussion should be a worthwhile reference as you continue to sort through these ideas. @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ #$&* If the average rate at which position changes with respect to clock time is 5 cm / second, and if the clock time changes by 10 seconds, by how much does the position change? answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): D=r * t D= 5cm/s * 10s D= 50 cm/s^2 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ #$&* You will be expected hereafter to know and apply, in a variety of contexts, the definition given in this question. You need to know this definition word for word. If you try to apply the definition without using all the words it is going to cost you time and it will very likely diminish your performance. Briefly explain how you will ensure that you remember this definition. answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): You must know the definition of the terms word for word and know how to apply it to different contexts, this will save you time and increase your performance @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ #$&* You are asked in this exercise to apply the definition, and given a general procedure for doing so. Briefly outline the procedure for applying this definition, and briefly explain how you will remember to apply this procedure. answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): in this procerdure rate of change of an object is defined by the change of its distance divided by its change of time. I use the formula D=r * t to find the solution and manipulate the formula to find the answer I am looking for. @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ #$&*