The following is a solution to the given problem.

Please compare the given solution with your solution and submit a self-critique of any error(s) in your solutions, and/or and additional questions or comments you might have.

Simply copy your posted document into a text editor and insert revisions, questions, and/or self-critiques, marking your insertions with ####. Submit using the Submit Work Form.

Among other things, this problem should illuminate the differences between a number of easily-confused idea, including

midpoint clock time,
change in clock time,
clock time,
time interval,
initial velocity,
final velocity,
average velocity,
change in velocity and
the meaning of the slope of a v vs. t graph.

What is the clock time at the midpoint of this interval?

The graph is a straight line, so the midpoint lies halfway between the two clock times, and halfway between the two velocities.

The quantity halfway between two values is the mean of the two values, obtained by adding the values and dividing by 2.

The midpoint of the time interval between (5 sec, 16 cm/s) to the point (13 sec, 40 cm/s) is the clock time midway between 5 sec and 13 sec

This midpont occurs at clock time (5 sec + 13 sec) / 2 = 9 sec.

What is the velocity at the midpoint of this interval?

The midpoint of the velocity between 16 cm/s and 40 cm/s occurs midway between the two velocities, at (16 cm/s + 40 cm/s) / 2 = 28 cm/s.

How far do you think the object travels during this interval?

Note that the duration of the time interval from t = 5 sec to t = 13 sec is 13 sec - 5 sec = 8 sec, which is not the same as the midpoint of the time interval.

Be sure you understand the differences among the terms clock time, time interval and midpoint clock time. It's easy to confuse some of these terms with others.
Symbolically these quantities are represented here as t (clock time), `dt (time interval) and t_mid (midpoint clock time).

In commonsense terms, we see that the object moves at average velocity 28 cm/s for 8 seconds, and so travels 28 cm/s * 8 s = 224 cm.

You should also understand this more formally, in terms of the relevant symbols and definitions:

If the graph is a straight line, the midpoint velocity will be the average velocity on the interval.

The average velocity is the average rate of change of position with respect to clock time, i.e.,

vAve = `ds / `dt

We want to find the displacement `ds and we now know vAve.

Knowing vAve, all we need in order to find the displacement `ds is the value of `dt, the duration of the time interval.

If we know `dt as well as vAve, we can rearrange the equation

vAve = `ds / `dt

to obtain

`ds = vAve * `dt

and simply substitute our values of vAve and `dt.

The time interval lasts from t = 5 sec to t = 13 sec; the duration of the interval is therefore

`t = t_f = t_0 = 13 s - 5 s = 8 s.

Thus we can find the displacement `ds:

`ds = vAve * `dt = 28 cm/s * 8 s = 224 cm.

By how much does the clock time change during this interval?

The change in clock time from t = 5 sec to t = 13 sec is `dt = 13 sec - 5 sec = 8 sec.

By how much does velocity change during this interval?

The change in velocity during the interval is from 16 cm/s to 40 cm/s, a change of 24 cm/s. This change occurs in the 8-sec duration of the interval.

What is the average rate of change of velocity with respect to clock time on this interval?

The average rate of change of velocity with respect to clock time is

ave rate =

change in velocity / change in clock time =

( 24 cm/s ) / (8 s) =

3 (cm/s) / s =

3 (cm/s) * (1/s) =

3 cm/s^2.

Be sure you understand the algebra of the units.

What is the rise of the graph between these points?

The rise of the graph during the interval is from 16 cm/s to 40 cm/s, a rise of 24 cm/s.

What is the run of the graph between these points?

This run of the graph is the 8-sec duration of the time interval, which runs from 5 sec to 13 sec.

What is the slope of the graph between these points?

The slope of the graph is

graph slope =

rise / run =

( 24 cm/s ) / (8 s) =

3 (cm/s) / s =

3 (cm/s) * (1/s) =

3 cm/s^2.

You should recognize this as the same calculation done above for the average rate of change of velocity with respect to clock time.

The average rate of change of velocity with respect to clock time on an interval is, by definition, the average acceleration on that interval.
The change in velocity corresponds to the rise of the v vs. t graph, the change in clock time to the run, so the slope between two points of the v vs. t graph represents the average acceleration.

What does the slope of the v vs. t graph tell you about the motion of the object during this interval?

The slope calculation is identical to the calculation of the average rate of change of velocity with respect to clock time. The slope therefore represents this average rate of change.

The following is a solution to the given problem.

Please compare the given solution with your solution and submit a self-critique of any error(s) in your solutions, and/or and additional questions or comments you might have.

Simply copy your posted document into a text editor and insert revisions, questions, and/or self-critiques, marking your insertions with ####. Submit using the Submit Work Form.

Among other things, this problem should illuminate the differences between a number of easily-confused idea, including

midpoint clock time,

change in clock time,

clock time,

time interval,

initial velocity,

final velocity,

average velocity,

change in velocity and

the meaning of the slope of a v vs. t graph.

What is the clock time at the midpoint of this interval?

The graph is a straight line, so the midpoint lies halfway between the two clock times, and halfway between the two velocities.

The quantity halfway between two values is the mean of the two values, obtained by adding the values and dividing by 2.

The midpoint of the time interval between (5 sec, 16 cm/s) to the point (13 sec, 40 cm/s) is the clock time midway between 5 sec and 13 sec

This midpont occurs at clock time (5 sec + 13 sec) / 2 = 9 sec.

What is the velocity at the midpoint of this interval?

The midpoint of the velocity between 16 cm/s and 40 cm/s occurs midway between the two velocities, at (16 cm/s + 40 cm/s) / 2 = 28 cm/s.

How far do you think the object travels during this interval?

Note that the duration of the time interval from t = 5 sec to t = 13 sec is 13 sec - 5 sec = 8 sec, which is not the same as the midpoint of the time interval.

Be sure you understand the differences among the terms clock time, time interval and midpoint clock time. It's easy to confuse some of these terms with others.

Symbolically these quantities are represented here as t (clock time), `dt (time interval) and t_mid (midpoint clock time).

In commonsense terms, we see that the object moves at average velocity 28 cm/s for 8 seconds, and so travels 28 cm/s * 8 s = 224 cm.

You should also understand this more formally, in terms of the relevant symbols and definitions:

If the graph is a straight line, the midpoint velocity will be the average velocity on the interval.

The average velocity is the average rate of change of position with respect to clock time, i.e.,

vAve = `ds / `dt

We want to find the displacement `ds and we now know vAve.

Knowing vAve, all we need in order to find the displacement `ds is the value of `dt, the duration of the time interval.

If we know `dt as well as vAve, we can rearrange the equation

vAve = `ds / `dt

to obtain

`ds = vAve * `dt

and simply substitute our values of vAve and `dt.

The time interval lasts from t = 5 sec to t = 13 sec; the duration of the interval is therefore

`t = t_f = t_0 = 13 s - 5 s = 8 s.

Thus we can find the displacement `ds:

`ds = vAve * `dt = 28 cm/s * 8 s = 224 cm.

By how much does the clock time change during this interval?

The change in clock time from t = 5 sec to t = 13 sec is `dt = 13 sec - 5 sec = 8 sec.

By how much does velocity change during this interval?

The change in velocity during the interval is from 16 cm/s to 40 cm/s, a change of 24 cm/s. This change occurs in the 8-sec duration of the interval.

What is the average rate of change of velocity with respect to clock time on this interval?

The average rate of change of velocity with respect to clock time is

ave rate =

change in velocity / change in clock time =

( 24 cm/s ) / (8 s) =

3 (cm/s) / s =

3 (cm/s) * (1/s) =

3 cm/s^2.

Be sure you understand the algebra of the units.

What is the rise of the graph between these points?

The rise of the graph during the interval is from 16 cm/s to 40 cm/s, a rise of 24 cm/s.

What is the run of the graph between these points?

This run of the graph is the 8-sec duration of the time interval, which runs from 5 sec to 13 sec.

What is the slope of the graph between these points?

The slope of the graph is

graph slope =

rise / run =

( 24 cm/s ) / (8 s) =

3 (cm/s) / s =

3 (cm/s) * (1/s) =

3 cm/s^2.

You should recognize this as the same calculation done above for the average rate of change of velocity with respect to clock time.

The average rate of change of velocity with respect to clock time on an interval is, by definition, the average acceleration on that interval.

The change in velocity corresponds to the rise of the v vs. t graph, the change in clock time to the run, so the slope between two points of the v vs. t graph represents the average acceleration.

What does the slope of the v vs. t graph tell you about the motion of the object during this interval?

The slope calculation is identical to the calculation of the average rate of change of velocity with respect to clock time. The slope therefore represents this average rate of change.