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.
A ball is moving at 10 cm/s when clock time is 4 seconds, and at 40 cm/s when clock time is 9 seconds.
Sketch a v vs. t graph and represent these two events by the points (4 sec, 10 cm/s) and (9 s, 40 cm/s).
Sketch a straight line segment between these points.
What are the rise, run and slope of this segment?
The rise is from 10 cm/s to 40 cm/s, which is a rise of (40 cm/s - 10 cm/s) = 30 cm/s.
-
Note that 10 cm/s and 40 cm/s have the same units and are therefore like terms. As like terms they can be subtracted, and their difference is therefore a like term with the same units.
-
Similarly the run is from 4 sec to 9 sec, a run of (9 s - 4 s) = 5 s.
-
Slope is therefore rise / run = (30 cm/s) / (5 s) = 6 cm/s^2.
What is the area of the graph beneath this segment?
The region beneath this segment forms a trapezoid, with 'altitudes' representing the initial and final velocities 10 cm/s and 40 cm/s.
-
The average 'altitude' of the trapezoid is therefore (10 cm/s + 40 cm/s) / 2 = 25 cm/s, and represent the average of the initial and final velocities.
-
Since the graph is a straight line, the average altitude therefore represents the average velocity on the interval.
-
The width of the trapezoid is equal to the 'run' between the two points, which as seen earlier is 5 s. This represents duration of the time interval.
-
The area of the trapezoid is equal to average altitude * width = 25 cm/s * 5 s = 125 cm.
-
This area represents the product of the average velocity and the duration of the time interval. The
area therefore represents the displacement of the object during the time interval.
University Physics Students: Note the solution to the area question
using integration, after the Miscellaneous Notes below.
Miscellaneous notes:
- When looking at the region beneath the graph, it can
be viewed as a triangle with altitude 30 cm/s and width 5 s, on top of a
rectangle with 'altitude' 10 cm/s and width 5 s. The areas are
respectively 75 cm and 50 cm, so the total area is 125 cm, the same as
that of the trapezoid. However for the purpose of interpretation
of area it's preferable to see the region as a single region, a
trapezoid with an average 'altitude' and a width, which are multiplied
to get the area.
- The terminology used here to describe the trapezoid
that occurs on the graph is not standard geometric terminology. To avoid
confusion it is helpful to think in terms of 'graph trapezoids', which
are the trapezoids that naturally arise in finding the area beneath a
segment of a graph, and 'geometric trapezoids'.
- A 'graph trapezoid' has two sides, typically
unequal, parallel to the 'vertical' axis, and one side running along
the 'horizontal' axis. The fourth side runs along the 'top' of
the trapezoid and has a slope defined by its 'rise / run'. The
two 'vertical' sides are here called 'altitudes'.
- A geometric trapezoid is a quadrilateral with two
parallel sides, which are called its 'bases', and an altitude equal
to the distance between the lines defined by the parallel sides.
Its area is equal to the product of the average of the two bases,
and its altitude.
- Technically the 'altitudes' of the 'graph
trapezoid' are the bases of the 'geometric trapezoid', and what is
here called the 'width' of the graph trapezoid is the altitude of
the 'geometric trapezoid'. However for the purpose of
understanding the graph and its interpretation, the terminology used
here is more intuitive.
University Physics Students: Solution to the area question using
integration:
The area beneath a graph, on an interval for which the graph lies above the
horizontal axis, can be found by integrating the function being graphed between
the endpoints of the interval. So if we can find the velocity v as a
function of clock time t, we can easily perform the integration.
-
Thus we need a function to integrate, but don't yet have one.
The graph is a straight line, so our function is linear.
The graph is of velocity vs. clock time. So we need a velocity vs. clock
time function which takes the given values at the given point.
-
We therefore need the equation of a straight line through two the two
given points. This equation will define our function.
We can obtain the function as follows:
A line through the point (4 sec, 10 cm/s) and (9 sec, 40 cm/s) has slope
6 cm/s^2.
A straight line from the point (4 sec, 10 cm/s) to (v, t) has slope (v -
10 cm/s) / (t - 4 sec).
If the point (v, t) lies on the same straght line as the line through
the given two points, these two slopes must be identical
so
-
(v - 10 cm/s) / (t - 4 sec) = 6 cm/s^2.
The equation of the straight line (which could also have been obtained
by the point-slope form of a straight line) is therefore
-
(v - 10 cm/s) = 6 cm/s^2 * (t - 4 s) .
Solving for v we have
-
v = 6 cm/s^2 * t - 14 cm/s.
Thus the velocity function is
-
v(t) = 6 cm/s^2 * t - 14 cm/s
This function, between t = 4 s and t = 9 s, represents the straight line
between our data points.
-
The area beneath the resulting graph is therefore the integral of
this function, with respect to t, from t = 4 s to t = 9 s.
v(t) = 6 cm/s^2 * t - 14 cm/s is a linear function of the form v(t) = m * t
+ b. The general antiderivative of such a function is 1/2 m * t^2 + b * t
+ c, where c can be any constant (in this case m = 6 cm/s^2 and b = -14 cm/s).
Any antiderivative can be used to evaluate the definite integral using the
Fundamental Theorem of Calculus.
We will use 1/2 m * t^2 + b as the form of the antiderivative.
For the present values of m and b we find that our antiderivative is
6 cm/s^2 * t^2 / 2 - 14 cm/s * t =
3 cm/s^2 * t^2 - 14 cm/s * t.
The definite integral from t = 4 s to t = 9 s is therefore the change in
the antiderivative:
definite integral = (antiderivative evaluated at t = 9) - (antiderivative
evaluated at t = 4) =
(3 cm/s^2 * (9 s)^2 - 14 cm/s * 9 s) - (3 cm/s^2 * (4 s)^2 - 14 cm/s
* 4 s) =
(243 cm - 126 cm) - (48 cm - 56 cm) =
117 cm - (-8 cm) =
125 cm.
This agrees with the solution obtained by figuring out the area of the
'graph trapezoid'.