In this video, we will learn how to
calculate the displacement or acceleration of a particle moving in a straight line
from its velocity–time graph.
We will begin by recalling the key
features of a velocity–time graph. The velocity of an object is its
speed in a particular direction. It can therefore take positive or
negative values. It is usually measured in meters
per second but could also be measured in kilometers per hour or miles per hour. A velocity–time graph, therefore,
shows the speed and direction an object travels over a specific period of time. When the velocity is measured in
meters per second, the time will be measured in seconds. Likewise, if the velocity was in
miles per hour or kilometers per hour, the time would be measured in hours.
The vertical or 𝑦-axis represents
velocity, and the horizontal axis represents time. Whilst the velocity can take
positive and negative values, the time can only take positive values. We will now look at how we can use
a velocity–time graph to calculate the acceleration and displacement of an
object. Let’s firstly consider acceleration
on a velocity–time graph. When our velocity is in meters per
second, acceleration is measured in meters per square second or meters per second
per second. This can be written in either of
these two ways. The acceleration is the slope or
gradient of the graph. And in this video, we will only be
dealing with straight-line graphs.
We can therefore calculate the
acceleration by dividing the change in velocity by the change in time. This is the same as the change in
𝑦 over the change in 𝑥 or the rise over the run. If we consider the straight line
shown, we can calculate the acceleration by creating a right-angled triangle. The change in velocity is labeled
𝑦, and the change in time is labeled 𝑥. The velocity has changed from two
meters per second to six meters per second, so we need to subtract two from six. The time has gone from zero to six
seconds. Six minus two is equal to four, and
six minus zero is equal to six. By dividing the numerator and
denominator by two, this fraction simplifies to two-thirds.
The acceleration of the body shown
is two-thirds meters per square second or two-thirds meters per second per
second. We are dividing a unit in meters
per second by a unit in seconds. In this example, our line has a
positive slope or gradient. This means that the object is
accelerating. If our line has a negative slope or
gradient, the object is decelerating. This means it will have a negative
acceleration. If we have a horizontal line on our
velocity–time graph, the acceleration is zero. This means that the object is
moving with a constant velocity. We will now look at how we can
calculate the displacement on a velocity–time graph.
The area between the line and the
𝑥-axis represents the displacement of the object. If the area is above the 𝑥-axis,
then the displacement is positive. However, if the area is below the
𝑥-axis, the displacement is negative. We can use our values for
displacement on a velocity–time graph to calculate the total distance traveled. In order to make it easier to
calculate the displacement, we can split our graph into triangles, rectangles, and
trapezoids or trapeziums. We will now look at some examples
of velocity–time graphs.
Use the velocity–time graph
below to determine the velocity after five seconds and the time period when the
velocity is five meters per second.
In any velocity–time graph, we
have the velocity on the vertical or 𝑦-axis and the time on the horizontal or
𝑥-axis. The first part of this question
wants us to work out the velocity after five seconds. We can draw a line vertically
upwards from five seconds. Once we hit the velocity–time
graph, we draw a horizontal line across to the 𝑦-axis. This will give us the velocity
after five seconds. This is halfway between two and
three meters per second. We can therefore conclude that
after five seconds, the velocity is 2.5 meters per second.
The second part of our question
asks us to find the time period when the velocity is five meters per second. This occurs during the
horizontal part of the graph. During this period, the object
has constant velocity. This means it has zero
acceleration. Whilst it is not relevant to
this question, when the line is sloping upwards from left to right, it has
positive acceleration, and when it is sloping downwards, it has negative
acceleration. We can see that the body is
traveling at five meters per second between two points. One of these is at 10 seconds,
and the other one is halfway between 20 and 25 seconds. 22.5 is halfway between 20 and
25. Therefore, the object is
traveling at five meters per second between 10 seconds and 22.5 seconds.
The correct answers are
therefore 2.5 meters per second and the inequality 𝑡 is greater than or equal
to 10 seconds and less than or equal to 22.5 seconds.
We will now consider some questions
where we need to calculate the acceleration or displacement of an object.
The given velocity–time graph
represents a particle moving in a straight line. Determine its displacement at
𝑡 equals two seconds.
The displacement of any
particle is its distance from the origin or start point. When dealing with a
velocity–time graph, we know that the displacement is the area between the graph
and the 𝑥-axis. In this question, we need to
calculate the displacement at 𝑡 equals two seconds. In order to do this, we draw a
vertical line from two on the 𝑥- or horizontal axis. This creates a right-angled
triangle. The displacement of the
particle will be the area of this triangle. In order to calculate the area
of any triangle, we multiply the base by the height and then divide by two. The base of our triangle is
two, and the height is 30. We need to multiply two by 30
and then divide by two. This is equal to 30.
Our units for velocity in this
question are centimeters per second, and our units for time are seconds. This means that the units for
displacement will be centimeters. The displacement of the
particle at 𝑡 equals two seconds is 30 centimeters.
Our next question involves a
velocity–time graph with both positive and negative velocities.
Given the velocity–time graph
of a particle moving in a straight line, determine the distance covered by the
particle within the time interval zero to eight.
We recall that in any
velocity–time graph, the displacement is the area between the graph and the
𝑥-axis. Displacement can be both
positive and negative. If the area is above the
𝑥-axis, the displacement is positive. And if it is below the 𝑥-axis,
the displacement is negative. In this question, however,
we’re asked for the distance. This means that we’ll take the
absolute value of the displacements and add them. The value for our distance must
be positive. As we are looking at the time
interval from zero to eight seconds, part of our graph is above the 𝑥-axis and
part of it is below.
To calculate the distance
covered by the particle, we will need to calculate the area of the triangle, the
area of the trapezoid and add our answers. Had we been calculating the
displacement, we would’ve subtracted the area of the trapezoid or trapezium from
the area of the triangle. The area of any triangle can be
calculated by multiplying the base by the height and then dividing by two. The base of our triangle is
one, and its height is five. We need to multiply these
numbers and then divide by two. Five divided by two is equal to
2.5. As our units for velocity were
meters per second and our units for time were seconds, our units for distance
and displacement in this question will be meters. The particle traveled a
distance of 2.5 meters from zero to one second.
In order to calculate the area
of a trapezoid or a trapezium, we add the lengths of the parallel sides, divide
by two, and multiply by the perpendicular height. The parallel sides of the
trapezium have lengths seven and four. The height of the trapezoid or
distance between the parallel sides is five. 11 divided by two is 5.5. And multiplying this by five
gives us 27.5. The distance covered between
one and eight seconds is 27.5 meters. Had we been looking for the
displacement here, this would be negative 27.5 as it is below the 𝑥-axis. We can calculate the total
distance covered by the particle by adding 2.5 and 27.5. This is equal to 30 meters.
The final question that we will
look at deals with acceleration on a velocity–time graph.
The figure shown is a
velocity–time graph for a body moving in a straight line. Determine the deceleration of
the body during the final section of its movement, given that it came to rest
100 seconds after it started moving.
We know that the acceleration
on a velocity–time graph is equal to its slope or gradient. We can therefore calculate the
acceleration at any point by dividing the change in velocity by the change in
time. In this question, our velocity
is measured in meters per second, and our time is measured in seconds. Dividing meters per second by
seconds gives us meters per second per second. This is written as meters per
second squared or meters per square second.
When the slope or gradient of
our graph is positive, the body will be accelerating. When the slope or gradient of
the graph is negative, the body will be decelerating. Finally, if a line on our
velocity–time graph is horizontal, the acceleration will be equal to zero and
the body will be traveling with a constant velocity. In this question, we’re
interested in the deceleration in the final section of the graph. This occurs between 90 seconds
and 100 seconds. The velocity during this time
period has decreased from 45 meters per second to zero meters per second. This means that our change in
velocity is zero minus 45.
As mentioned, the time period
goes from 90 seconds to 100 seconds. Therefore, the change in time
is 100 minus 90. This can be simplified to
negative 45 over 10. When dividing by 10, we move
all our digits one place to the right. And dividing a negative number
by a positive gives a negative answer. The acceleration of the body is
therefore negative 4.5 meters per square second. The deceleration will be the
absolute value or modulus of this. This is equal to 4.5 meters per
We will now summarize the key
points of this video. In this video, we looked at how we
could use a velocity–time graph to calculate the displacement and acceleration of a
particle. The acceleration of a particle is
equal to the slope or gradient of the line. We found that this could be
positive, negative, or zero. A negative acceleration is also
known as a deceleration. This occurs when the line slopes
downwards. When our line is horizontal,
acceleration is equal to zero. This means the body is moving with
a constant velocity. We’re able to calculate the
acceleration at any point on our graph by dividing the change in velocity by the
change in time. This is the same as the rise over
We also found that the displacement
is the area between the graph and the 𝑥-axis. If the area is above the 𝑥-axis,
the displacement is positive, whereas if it is below the 𝑥-axis, the displacement
is negative. We also identified the difference
between the displacement and distance as displacement is distance with a
direction. Whilst we only looked at
straight-line graphs in this video, it is also possible to estimate the acceleration
and displacement of a particle on a curved velocity–time graph.