This is the first part of a 2-part video series on velocity-time graphs. We're not
going to get into the second part this unit because this unit is about constant velocity,
and so this first part is only going to cover graphs of constant velocity on velocity-time
graphs specifically, So velocity-time graphs show an object's velocity on the y-axis and the time
on the x-axis. They're used to show complicated motion on a simple picture. So as an example,
if this person were to start walking as the time moves, I can see that they have a velocity of 1,
and then a velocity of 2, and then a velocity of 0, and then a velocity of -1, and so that's
how that would be represented on the graph. I'm going to play that back once more just so you
can see it happening again. This person begins to move forward with a velocity of 1 for 30 seconds,
and then they have a velocity of 2 for 2 seconds, and then they're stopped with a velocity of 0,
and then they have a velocity of -1 for 4 seconds. So that's what the velocity versus time graph of
their motion would look like. So something you may have noticed here is that the y-intercept of
the velocity-time graph is the starting velocity of the object. They started with a velocity of 1
meter per second at a time equals 0. So that's what the y-intercept of a velocity-time graph
tells you. This next part can confuse students when they first see it but after some practice
I find that it becomes pretty easy. So the area under the curve of a velocity-time graph is the
area between the graph's line and the x-axis, and on a velocity-time graph specifically the
area under the curve equals the displacement of the object over that time. So I'll show you
what I mean by that. This is the area under the curve for the first part of the graph. It's the
area between the graph's line and the x-axis specifically, all the area that that takes up,
and so I can see here that that full area is the velocity times the time because the area of a
rectangle is equal to the base times the height. And because this is the velocity times the time,
that I remember that velocity times time is equal to displacement. So this area is going to be equal
to the displacement of the object, and when I multiply that out I can see that the area is
3 meters per second because that's the height of the velocity times the 5 seconds of time,
which is equal to 15 meters. So that area is equal to 15 meters. So in that time of 5 seconds the
object traveled positive 15 meters. This next area I can see is equal to positive 3 meters.
That's pretty easy, it's just 1 meter per second times the 3 seconds that that travels for. This
next part doesn't have any area under the curve between the line and the x-axis so it travels 0
meters in that time. In the last part I can see that this area is actually below the x-axis,
so it's still between the line and the x-axis, it's just negative, so this is actually going
to be a negative displacement of the object, because the velocity is negative it's going to
be negative 2 times 2 which is equal to negative 4 meters. So the total displacement that this
object traveled in this time is 15 plus 3 plus 0 minus 4 which is equal to 14 meters. So we can
find the displacement of the object from the area under the curve of a velocity-time graph. We can
try that with our original velocity-time graph just to see how that works. So I'm gonna have
this person move in this way. So I can see here that the area under the curve for that first part
is 1 meter per second times 3 seconds, which is equal to positive 3 meters, and I can see
that this person did actually move positive 3 meters. Next, they're going to move at 2 meters
per second for 2 seconds, and I can see that that area is 2 meters per second times 2 seconds which
is equal to 4 meters, and I can see that they did move another 4 meters there, and altogether
they've moved 7 meters. The next part they're going to be stopped, and then they're going to
go backwards at negative 1 meters per second for 4 seconds, and I can see here that their
displacement is going to be negative 1 meters per second times 4 seconds which is negative 4,
and again I can see that traveled negative 4 and so altogether their total displacement is 3 plus
4 plus 0 plus -4 which is equal to positive 3. So that fits our understanding of the displacement.