
Kinematics with Graphs
Since you are not allowed to use calculators, SAT II Physics
places a heavy emphasis on qualitative problems. A common way of
testing kinematics qualitatively is to present you with a graph
plotting position vs. time, velocity vs. time, or acceleration vs.
time and to ask you questions about the motion of the object represented
by the graph. Because SAT II Physics is entirely made up of multiplechoice
questions, you won’t need to know how to draw graphs; you’ll just
have to interpret the data presented in them.
Knowing how to read such graphs quickly and accurately
will not only help you solve problems of this sort, it will also
help you visualize the oftenabstract realm of kinematic equations.
In the examples that follow, we will examine the movement of an
ant running back and forth along a line.
Position vs. Time Graphs
Position vs. time graphs give you an easy and obvious
way of determining an object’s displacement at any given time, and
a subtler way of determining that object’s velocity at any given
time. Let’s put these concepts into practice by looking at the following
graph charting the movements of our friendly ant.
Any point on this graph gives us the position of the ant
at a particular moment in time. For instance, the point at (2,–2) tells
us that, two seconds after it started moving, the ant was two centimeters
to the left of its starting position, and the point at (3,1) tells
us that, three seconds after it started moving, the ant is one centimeter
to the right of its starting position.
Let’s read what the graph can tell us about the ant’s
movements. For the first two seconds, the ant is moving to the left.
Then, in the next second, it reverses its direction and moves quickly
to y = 1. The ant then stays still
at y = 1 for three seconds before
it turns left again and moves back to where it started. Note how
concisely the graph displays all this information.
Calculating Velocity
We know the ant’s displacement, and we know how long it
takes to move from place to place. Armed with this information,
we should also be able to determine the ant’s velocity, since velocity
measures the rate of change of displacement over time. If displacement
is given here by the vector y,
then the velocity of the ant is
If you recall, the slope of a graph is a measure of rise
over run; that is, the amount of change in the y direction
divided by the amount of change in the x direction.
In our graph, is the change in the y direction
and is the change in the x direction,
so v is a measure of
the slope of the graph. For any position vs. time graph,
the velocity at time t is equal to the slope of
the line at t. In a graph made up of straight
lines, like the one above, we can easily calculate the slope at
each point on the graph, and hence know the instantaneous velocity
at any given time.
We can tell that the ant has a velocity of zero from t = 3 to t = 6,
because the slope of the line at these points is zero. We can also
tell that the ant is cruising along at the fastest speed between t = 2 and t = 3,
because the position vs. time graph is steepest between these points.
Calculating the ant’s average velocity during this time interval
is a simple matter of dividing rise by run, as we’ve learned in
math class.
Average Velocity
How about the average velocity between t = 0 and t = 3?
It’s actually easier to sort this out with a graph in front of us,
because it’s easy to see the displacement at t = 0 and t = 3,
and so that we don’t confuse displacement and distance.
Average Speed
Although the total displacement in the first three seconds
is one centimeter to the right, the total distance traveled is two
centimeters to the left, and then three centimeters to the right, for
a grand total of five centimeters. Thus, the average speed is not
the same as the average velocity of the ant. Once we’ve calculated
the total distance traveled by the ant, though, calculating its
average speed is not difficult:
Curved Position vs. Time Graphs
This is all well and good, but how do you calculate the
velocity of a curved position vs. time graph? Well, the bad news
is that you’d need calculus. The good news is that SAT II Physics
doesn’t expect you to use calculus, so if you are given a curved
position vs. time graph, you will only be asked qualitative questions
and won’t be expected to make any calculations. A few points on
the graph will probably be labeled, and you will have to identify which
point has the greatest or least velocity. Remember, the point with
the greatest slope has the greatest velocity, and the point with
the least slope has the least velocity. The turning points of the
graph, the tops of the “hills” and the bottoms of the “valleys”
where the slope is zero, have zero velocity.
In this graph, for example, the velocity is zero at points A and C,
greatest at point D, and smallest
at point B. The velocity at point B is
smallest because the slope at that point is negative. Because velocity
is a vector quantity, the velocity at B would
be a large negative number. However, the speed at B is
greater even than the speed at D:
speed is a scalar quantity, and so it is always positive. The slope
at B is even steeper than at D,
so the speed is greatest at B.
Velocity vs. Time Graphs
Velocity vs. time graphs are the most eloquent kind of
graph we’ll be looking at here. They tell us very directly what
the velocity of an object is at any given time, and they provide subtle
means for determining both the position and acceleration of the
same object over time. The “object” whose velocity is graphed below
is our everindustrious ant, a little later in the day.
We can learn two things about the ant’s velocity by a
quick glance at the graph. First, we can tell exactly how fast it
is going at any given time. For instance, we can see that, two seconds
after it started to move, the ant is moving at 2 cm/s.
Second, we can tell in which direction the ant is moving. From t =
0 to t = 4, the velocity is
positive, meaning that the ant is moving to the right. From t =
4 to t = 7, the velocity is
negative, meaning that the ant is moving to the left.
Calculating Acceleration
We can calculate acceleration on a velocity vs.
time graph in the same way that we calculate velocity on a position
vs. time graph. Acceleration is the rate of change of the velocity
vector, , which expresses itself
as the slope of the velocity vs. time graph. For a velocity vs. time
graph, the acceleration at time t is equal
to the slope of the line at t.
What is the acceleration of our ant at t = 2.5 and t = 4? Looking
quickly at the graph, we see that the slope of the line at t = 2.5 is
zero and hence the acceleration is likewise zero. The slope of the
graph between t = 3 and t = 5 is
constant, so we can calculate the acceleration at t = 4 by
calculating the average acceleration between t = 3 and t = 5:
The minus sign tells us that acceleration is in the leftward
direction, since we’ve defined the ycoordinates
in such a way that right is positive and left is negative. At t = 3,
the ant is moving to the right at 2 cm/s, so a leftward
acceleration means that the ant begins to slow down. Looking at
the graph, we can see that the ant comes to a stop at t = 4,
and then begins accelerating to the right.
Calculating Displacement
Velocity vs. time graphs can also tell us about an object’s
displacement. Because velocity is a measure of displacement over
time, we can infer that:
Graphically, this means that the displacement
in a given time interval is equal to the area under the graph during
that same time interval. If the graph is above the taxis,
then the positive displacement is the area between the graph and
the taxis. If the graph is below
the taxis, then the displacement
is negative, and is the area between the graph and the taxis. Let’s
look at two examples to make this rule clearer.
First, what is the ant’s displacement between t =
2 and t = 3? Because the velocity
is constant during this time interval, the area between the graph
and the taxis is a rectangle of width 1 and
height 2.
The displacement between t = 2 and t =
3 is the area of this rectangle, which is 1 cm/ss
= 2 cm to the right.
Next, consider the ant’s displacement between t =
3 and t = 5. This portion of
the graph gives us two triangles, one above the taxis
and one below the taxis.
Both triangles have an area of ^{1}
/_{2} (1 s)(2 cm/s)
= 1 cm. However, the first triangle is above the taxis,
meaning that displacement is positive, and hence to the right, while
the second triangle is below the taxis,
meaning that displacement is negative, and hence to the left. The
total displacement between t = 3 and t =
5 is:
In other words, at t = 5,
the ant is in the same place as it was at t =
3.
Curved Velocity vs. Time Graphs
As with position vs. time graphs, velocity vs. time graphs
may also be curved. Remember that regions with a steep slope indicate
rapid acceleration or deceleration, regions with a gentle slope
indicate small acceleration or deceleration, and the turning points
have zero acceleration.
Acceleration vs. Time Graphs
After looking at position vs. time graphs and velocity
vs. time graphs, acceleration vs. time graphs should not be threatening.
Let’s look at the acceleration of our ant at another point in its
dizzy day.
Acceleration vs. time graphs give us information about
acceleration and about velocity. SAT II Physics generally sticks
to problems that involve a constant acceleration. In this graph,
the ant is accelerating at 1 m/s^{2} from t =
2 to t = 5 and is not accelerating
between t = 6 and t =
7; that is, between t = 6 and t =
7 the ant’s velocity is constant.
Calculating Change in Velocity
Acceleration vs. time graphs tell us about an object’s
velocity in the same way that velocity vs. time graphs tell us about
an object’s displacement. The change in velocity in a given
time interval is equal to the area under the graph during that same
time interval. Be careful: the area between the graph and
the taxis gives the change in
velocity, not the final velocity or average velocity over a given
time period.
What is the ant’s change in velocity between t =
2 and t = 5? Because the acceleration
is constant during this time interval, the area between the graph
and the taxis is a rectangle of height 1 and
length 3.
The area of the shaded region, and consequently the change
in velocity during this time interval, is 1 cm/s^{2} · 3 s
= 3 cm/s to the right. This doesn’t mean that the velocity
at t = 5 is 3 cm/s; it
simply means that the velocity is 3 cm/s greater than
it was at t = 2. Since we have not
been given the velocity at t = 2,
we can’t immediately say what the velocity is at t =
5.
Summary of Rules for Reading Graphs
You may have trouble recalling when to look for the slope
and when to look for the area under the graph. Here are a couple
handy rules of thumb:
 The slope on a given graph is equivalent to the quantity we get by dividing the yaxis by the xaxis. For instance, the yaxis of a position vs. time graph gives us displacement, and the xaxis gives us time. Displacement divided by time gives us velocity, which is what the slope of a position vs. time graph represents.
 The area under a given graph is equivalent to the quantity we get by multiplying the xaxis and the yaxis. For instance, the yaxis of an acceleration vs. time graph gives us acceleration, and the xaxis gives us time. Acceleration multiplied by time gives us the change in velocity, which is what the area between the graph and the xaxis represents.
We can summarize what we know about graphs in a table:
