Video Transcript
The velocity–time graph shows the
change in the velocity of a person walking in the time interval from 𝑡 equals zero
seconds to 𝑡 equals six seconds. The person’s starting point is
their position when 𝑡 equals zero seconds. What is the person’s displacement
from their starting point after three seconds?
Here, we are given a velocity–time
graph that shows the change in the velocity of a person walking in a time interval
from zero to six seconds. We are asked to use this graph to
figure out what the person’s displacement is during the first three seconds of their
movement. Let’s first refresh our memories
about velocity–time graphs and how we can use them to find the displacement of an
object.
Looking at the graph, we can see
that the vertical axis shows the object’s velocity and the horizontal axis shows the
time that the object has traveled for. When an object is moving with a
constant velocity, then when its motion is plotted on a velocity–time graph, the
height of the line on the velocity axis won’t change for increasing time values. Therefore, motion at constant
velocity is represented by a straight horizontal line on a velocity–time graph.
Notice that the first two seconds
of movement are at a constant velocity, as well as the last two seconds. A straight, diagonal line on a
velocity–time graph represents motion with a constant acceleration. A positive slope shows the velocity
increasing as time passes and so corresponds to a positive acceleration. Similarly, a negative slope shows
decreasing velocity and so represents a negative acceleration. We can see that the person
undergoes negative acceleration between the time of two seconds and four
seconds.
Let’s now recall how we can find
the displacement of an object from a velocity–time graph. The displacement is given by the
area between the line on the graph representing the object’s motion and the
horizontal axis. When a velocity–time graph is made
up of different sections of motions, we can calculate the area for each section
separately and then add those areas together to find the total displacement.
For graphs such as this one,
containing both positive and negative velocity values, we need to be careful. When the velocity is positive and
the line is above the horizontal axis, the displacement is positive, while when the
velocity is negative and the line is below the horizontal axis, the displacement is
negative.
The question is asking us to find
the person’s displacement during the first three seconds of motion. So we need to find the area under
the line between zero seconds and three seconds. That is, we need to find the area
that we have shaded. Happily, we can see that the motion
during this time interval is all at a positive velocity. And so we don’t need to worry about
any negative displacements.
Looking at the graph, we can break
up this time into two parts: one that we’ve highlighted in pink between zero and two
seconds with constant velocity and one highlighted in orange between two and three
seconds with constant acceleration. Let’s now clear some space on
screen so that we can calculate these two areas.
Area one, that is the area under
the graph between zero and two seconds, is a rectangle. Recall that we can find the area of
a rectangle by multiplying its height by its width. Area one is a rectangle with a
height of three meters per second and a width of two seconds. We’ve labeled the height as ℎ one
and the width as 𝑤 one. Then area one must be equal to ℎ
one multiplied by 𝑤 one, which is three meters per second times two seconds. Canceling the units of seconds and
per second, we then find that area one is equal to six meters. So in the first two seconds, the
person here has moved six meters from their starting point.
Now turning our attention to area
two, we can see that this is a right-angled triangle. Recall that the area of a triangle
is equal to one-half multiplied by the base multiplied by the height. We can see that area two has a
height of three meters per seconds and a base of three seconds minus two seconds,
which is one second. We’ve labeled the height as ℎ two
and the base as 𝑏 two. Then, area two must be equal to a
half multiplied by ℎ two multiplied by 𝑏 two, which is a half times three meters
per second times one second. This works out as 1.5 meters.
So during the time between two
seconds and three seconds, the person becomes displaced by a further 1.5 meters from
their starting point. In order to find the total
displacement, we just need to add these two areas together. That is, the displacement from the
starting point after three seconds is equal to area one plus area two. Substituting in the values for
these areas, we have that the displacement is equal to six meters plus 1.5
meters. This works out as 7.5 meters.
Our answer then is that the
person’s displacement from their starting point after three seconds is equal to 7.5
meters.
