VELOCITY-TIME
GRAPHS: UNIFORM AND
NON-UNIFORM MOTION
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KINEMATICS: Lesson 3
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The shape of the velocity-time graph reveals
whether the object is at rest, moving at constant
speed, speeding up, or slowing down.
Suppose an airplane has a cruising altitude of
10 600 m and travels at a constant velocity of
900 km/h [E] for 5.0 h.
DATA TABLE
TABLE & GRAPH
1. SLOPE OF A VELOCITY-TIME
GRAPH
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Just as the slope of a position-time graph reveals the
rate at which position changes (velocity), the slope of a
velocity-time graph reveals the rate at which velocity
changes (acceleration).
SLOPES OF VELOCITY-TIME GRAPHS
The slope of a velocity-time graph represents the acceleration of the
object.
Non-uniform Motion – riding a quad
Velocity is not constant.
a. Definition of ACCELERATION
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An object whose velocity changes (in this case,
increases) over a time interval is undergoing
acceleration, represented by the variable a.
Acceleration is a vector quantity.
It is also called non-uniform motion because
the object’s speed or direction is changing.
When velocity decreases it is undergoing negative
acceleration, also represented by the variable a.
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Let’s determine the slope of the plane’s velocity time
graph. Designate east as the positive direction:
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Notice the units of the slope of the velocity-time
graph: km/h2.
These units are units of acceleration.
Because the plane is moving at a constant
velocity, its acceleration is zero.
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In general, you can recognize acceleration values
by their units.
In physics, the standard units for acceleration
are metres per second per second, which is
generally abbreviated to m/s2. (read metres per
second squared).
b. NEGATIVE ACCELERATION
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What is the meaning of negative acceleration?
When interpreting the sign of acceleration, you
need to compare it to the sign of velocity.
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When the directions (signs) of velocity and
acceleration are the same (positive or negative),
the object is speeding up.
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When velocity and acceleration have opposite
directions (signs), the object slows down.
2. AREA UNDER A VELOCITY-TIME
GRAPH
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The area under a velocity-time graph represents the
objects displacement.
Unit analysis indicates
that the area under a
velocity-time graph
equals displacement.
Multiplying l x w of each
rectangle gives us
For eastward displacement, the area is above the time axis, so it is
positive. For westward displacement, the area is below the time axis,
so it is negative.
EXAMPLE 1.6
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To calculate displacement we must find the sum of the three
areas (A, B and C) under the velocity-time graph.
Designate east as the positive direction (above the horizontal
axis) .
Convert minutes to seconds.
Let’s do this together!
CALCULATIONS FOR PART A:
b. To find acceleration, find the slope of each section of the graph.
Region C is a horizontal line so its slope is also zero.
3. AVERAGE VELOCITY ON A
VELOCITY – TIME GRAPH
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To calculate average velocity, determine the
area under the velocity-time graph and divide it
by the total time.
To find the average velocity given different
displacements over different time intervals,
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add the total displacement and divide by the total
time.
EXAMPLE 1.7
EXAMPLE 1.8
A bird starts flying south. Its motion is described in the velocity-time
graph in Figure 1.46.
From the graph, determine:
(a) whether acceleration is
positive, negative, or zero
for each section
(b) when the bird changes
direction.
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Acceleration is the slope of each section of the graph.
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A: Negative
Final velocity is more negative than the initial velocity, as the bird is
speeding up in the south direction. So the slope of this line is negative.
The bird’s acceleration is negative.
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B: Zero
Acceleration is zero because the slope is zero (the graph is a horizontal
line.)
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C: Negative
Acceleration is negative because the slope of the line is negative (as in
section A).
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D: Zero
Acceleration is zero because the slope of the line is zero (as in section B).
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E: Positive
Final velocity is positive because the bird is now flying north. So the slope
of this line is positive. The bird’s acceleration is positive.
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F: Zero
Acceleration is zero because the slope of the line is zero.
4. INSTANTANEOUS VELOCITY
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Instantaneous velocity is the velocity of an
object taken at a specific time, an instant.
Earlier in this section we learned that to
determine the velocity of an object from a
position-time graph you calculate the slope of the
graph.
How can you obtain the slope of a curve?
INSTANTANEOUS VELOCITY
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To determine the velocity (and therefore slope) of
an object at any instant, physicists use tangents.
A tangent is a straight line that touches a curve
at only one point.
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Each tangent on a curve has a unique slope,
which represents the velocity at that instant.
In order for the object to be at that position, at
that time, it must have an instantaneous velocity
equal to the slope of the tangent at that point.
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You can now create a new table using the slopes of the
position-time graphs.
This velocity-time graph represents an object
undergoing uniformly accelerated motion.
SLOPE OF A VELOCITY-TIME GRAPH
SLOPE OF A VELOCITY-TIME GRAPH
An acceleration-time graph for an object undergoing uniformly
accelerated motion is straight line with zero slope.
EXAMPLE 1.5
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This ATV is undergoing non-uniform motion. It is
accelerating, in this case, slowing down.
 What would the position-time graph look like?
 What would the velocity-time graph look like?
 What would the acceleration-time graph look like?
EXAMPLE 1.5
How do we determine the acceleration?
The acceleration of the ATV is 2.8 m/s2. Because the forward direction
was designated as positive, the negative sign means that the direction of
acceleration is backward.
SUMMARY
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Velocity-time graphs are very useful. They provide
the following information:
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The velocity-time graph gives you instantaneous velocity
values at given times.
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Calculating the slope of a velocity-time graph gives you an
object’s acceleration.
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The area under a velocity-time graph gives you the
object’s displacement.
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You can determine the average velocity of an object over a
time interval from a velocity-time graph.
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Homework from Textbook:
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Check and Reflect
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Page 30: Q # 1 – 4
Practice Problems