### Video Transcript

In this video, we will be looking
at how we can analyse an object’s motion when it is accelerating. Whilst also considering the
distance moved by the object as it accelerates, as well as the time taken for the
acceleration to occur.

So let’s start by imagining that we
have an object here, which we can say is initially moving towards the right at a
velocity 𝑢. But then, right at this point in
time, when the object is at this location, it starts accelerating. And let’s say that it starts
accelerating in the same direction as its motion. So let’s say it has a constant
acceleration, 𝑎.

Now because of this, sometime
later, the object will be at, let’s say, this location here. And because it’s been accelerating,
it will no longer be travelling at a velocity 𝑢. Instead, it will be travelling at a
different velocity. Let’s call this velocity 𝑣.

So to recap, what’s happening is
that we started with an object initially travelling at a velocity 𝑢, which then
accelerated at a constant rate, 𝑎, in the same direction as its velocity. And that object then ends up at
this location here, with a new velocity 𝑣. So essentially, the object has sped
up. Now let’s also say that the object
travels a certain distance, which we will call 𝑠, as it accelerates. And let’s also say that, at the
beginning of the object’s motion, we started a stopwatch. So the stopwatch read zero. And at the end of its trajectory,
it read a time 𝑡. In other words then, the object
accelerated over a time interval 𝑡. Or the amount of time taken for the
object to accelerate was 𝑡.

Well, for a situation like this,
there’s an equation that we can find that links all of the quantities that we’ve
labelled on this diagram. But in order to find this
relationship, what we’re going to do is to plot a velocity-time graph. So let’s draw our axes, and let’s
label the velocity on the vertical axis and time on the horizontal axis.

Now we’re going to be representing
the motion of this object on the velocity-time graph. So if we first label the origin on
our graph, we can then see that, at the beginning of the object’s trajectory, in
other words, at time is equal to zero, the object started with a certain
velocity. That velocity was 𝑢. So let’s plot that on the
graph. So at a time of zero, which means
it’s going to be on the vertical axis, we plot a velocity, let’s say, here, which is
𝑢. And then we know that the object
accelerated at a constant rate, which is 𝑎. We’ll come back to that in a
second.

But we know the object finished
over here at a time 𝑡 with a velocity 𝑣. So let’s say that, on the
horizontal axis, this is a time 𝑡. So at that time, we need to plot
the object’s velocity as being 𝑣. Now because the object is
accelerating in the same direction as its motion, 𝑣 is going to be larger than
𝑢. And this is because if we
accelerate in the same direction as our motion, then our velocity will increase in
that direction.

And so let’s say that, somewhere
here, it’s velocity 𝑣 on the vertical axis. And we know that the velocity 𝑣 is
attained at a time 𝑡. So to plot our point, what we need
to do is to go up from the time 𝑡 until we reach a point that represents the
velocity 𝑣 along the vertical axis. And so the point that we’re
plotting is here. That’s the final velocity of the
object at a time 𝑡.

And remember from earlier, at time
zero, so right at the beginning, the object’s velocity was 𝑢. So we should also place a cross
here to represent the initial point. And so, in other words, on our
velocity-time graph, the object starts up here and ends up here. Now we can draw a line between
these two points that will represent the velocity of the object over its entire
trajectory. In other words, if we draw a
straight line between the two points, then if we look at any point along that line —
so let’s say this point here — then that will allow us to work out the velocity of
the object at a certain point in time along its trajectory.

But then why is it that we’ve drawn
a straight line between the two points that we plotted, this point and this point
here? Why is it not a curved line? Well, it’s because, like we said
earlier, the acceleration of the object is constant. And what that essentially means is
that, for every unit of time that we move along, the velocity of the object
increases by the same amount. It’s speeding up at the same rate,
and so the acceleration is constant.

And another way to think about this
is to realise that the slope of a velocity-time graph gives the acceleration of the
object in question. And since we’re talking about a
constant acceleration, the slope must be constant as well, which means that this
line must be a straight line. Now on this graph, we’ve plotted
the initial velocity 𝑢, the final velocity 𝑣, and the total time of the object’s
motion, which is 𝑡. And as well as this, we can deduce
the acceleration of the object by finding the slope of the straight line. So the only thing that we haven’t
yet talked about is the distance covered by the object as it accelerates from here
going to here.

Well, we can actually work this out
from the graph as well. Let’s recall that the area under a
velocity-time graph will give us the displacement of the object in question. So what we’re trying to do is to
find the area underneath this line and above the horizontal axis. And this area will be the
displacement of the object that we’re talking about.

Now there are a couple of things to
note here. Firstly of course, we want to find
the area between the times at which we’re considering the object, zero and 𝑡. And so we want to find the area
between here and here. In other words then, this is the
area that we’re looking for. And secondly, we’ve said that the
area will give us the displacement of the object, not necessarily the distance
covered by the object. And this is an important point to
make. Because, remember, the displacement
of an object is the shortest distance between its start and finish point, in other
words, the shortest distance between here and here.

But then the shortest distance
between two points is a straight line. And when an object accelerates at a
constant rate in the same direction, then it will move in a straight line. So in this case, the distance that
we’ve labelled 𝑠 is indeed the area under the curve.

Now things can get a little bit
more hairy when the acceleration is in the opposite direction to the object’s
initial velocity, meaning that the object is actually decelerating. So what we mean is that if we
started with our object and it was initially moving to the right, but the
acceleration of the object was towards the left. Then what would happen to the
object is that it starts by moving out right. But it gets slower and slower until
eventually it stops. And then it doubles back on itself
because it continues to accelerate in the leftward direction. So it moves further and further
left.

Well, in that case, if we calculate
the area under our velocity-time graph, which will look very different to the one
that we’ve drawn here, and if we remember to account for the vector nature of
displacement. In that any area above the
horizontal axis is positive and any area below the horizontal axis is negative. Then this will give us the
displacement of the object. In other words, the shortest
distance between its start point and its finish point. And that’s this distance here, not
the total distance moved by the object. We would get this though if we just
did a normal area calculation for both areas above and below the horizontal axis and
simply added them together. And so that’s important to
remember. However, that’s a more complicated
example. So let’s not worry about that too
much right now.

Instead, let’s go about trying to
find the area under this graph, which represents an object for which its initial
velocity and its constant acceleration are in the same direction. Now there are a couple of different
ways to find the area under our graph. But a really convenient way is to
split it up into two areas. The first area is the rectangular
one underneath the dotted line. And the second is the triangular
one above the dotted line. Let’s call these areas area one and
area two, respectively.

Now of course, as we’ve said
already, the total area underneath the line represents the displacement of the
object 𝑠. And so we can say that the total
area, 𝑠, is equal to area one plus area two. So let’s go about calculating
firstly what area one is.

Now area one is a rectangle with a
width which is 𝑡, because the width is calculated by 𝑡 minus zero, which is simply
𝑡. And it has a height 𝑢, because
once again this length is calculated as 𝑢 minus zero. And so we can say that area one,
which is the area of the rectangle, is equal to the height of the rectangle
multiplied by the width of the rectangle. Because that’s how we find the area
of a rectangle, the height multiplied by the width.

And then we can move on to
calculating area number two, which is the area of a triangle. Now we can recall that the area of
a triangle is equal to half multiplied by the base multiplied by the height. Where in this case the base of the
triangle is this length here, and the height of the triangle is this length
here.

Now as we can see from the diagram,
the length of the triangle’s base is the same as the rectangle’s width. And so we know that the base of the
triangle is 𝑡. And then we know that the height of
the triangle is this distance here. In other words, that’s 𝑣 minus
𝑢. And so let’s label that height 𝑣
minus 𝑢. We can label it on the right-hand
side as well to make life easier for ourselves in terms of visualising the triangle
itself.

And so, at this point, we can say
that area number two, the area of the triangle, is equal to half multiplied by the
base, which we know is 𝑡, multiplied by the height, which we’ve just calculated to
be 𝑣 minus 𝑢. Now at this point, let’s realise
something. 𝑣 minus 𝑢 is simply the change in
velocity of the object over its entire trajectory, because, remember, the object
started with a velocity 𝑢 and finished with a velocity 𝑣. So we can say that the final
velocity minus the initial velocity is the same thing as the change in velocity,
which we will represent as Δ𝑣.

So now what we’re going to do is to
write this expression for area number two in terms of the acceleration of the object
because we know that this acceleration is constant. And to do this, we need to recall
that the acceleration of an object if it’s a constant acceleration is defined as the
change in velocity of the object divided by the time taken for that acceleration to
occur. Now in this case, we know that the
object is accelerating from here to here. And it takes a time 𝑡 to do
so. And so all of these quantities
check out. 𝑎 is the acceleration, Δ𝑣 is the
change in velocity, 𝑣 minus 𝑢, and the time taken is time 𝑡.

So what we’re going to do here is
to rearrange this equation by multiplying both sides by 𝑡. This way, the 𝑡 on the right-hand
side cancels. And what we’re left with is 𝑎
multiplied by 𝑡 on the left-hand side and Δ𝑣 on the right-hand side. This means that we can take our
Δ𝑣, which happens to be 𝑣 minus 𝑢, and replace it with 𝑎𝑡 because they’re the
same thing, as we’ve seen from this equation. And so what we end up with is that
area number two, the area of the triangle, is equal to half multiplied by 𝑡
multiplied by 𝑎𝑡. At which point we can simplify this
further because we can see that we’ve got two powers of 𝑡 in this expression. And so we can write this as half
multiplied by 𝑎 multiplied by 𝑡 squared.

So what we’ve done so far then is
to take this expression on the right-hand side and write it in terms of the
acceleration 𝑎 and the time over which it accelerates 𝑡. As opposed to writing it in terms
of its initial and final velocity, which looks a bit messier and is also not written
then in terms of the acceleration 𝑎.

And so the final step is to put
everything together into this equation here. We can say finally that the total
area underneath this line between time zero and time 𝑡 is given by the following
equation. The area representing the total
displacement of the object is equal to area number one, the area of the rectangle,
which is 𝑢𝑡, plus area number two, the area of the triangle, which is half 𝑎𝑡
squared.

And at this point, we’ve come up
with a very important equation, which happens to be one of the kinematic
equations. This equation comes in very handy
when we’ve got an object accelerating at a constant rate. And we know three out of the four
quantities mentioned in this equation. Well, these quantities are the
displacement of the object, the initial velocity, the time taken for acceleration,
and the acceleration itself. If we either already know or can
work out three out of the four quantities in this equation, then we can use the
equation to work out the fourth quantity.

But of course, it’s important to
remember that this equation only works when we’re dealing with a constant
acceleration in a straight line. Because if that wasn’t the case,
then our line here would not be a straight line. If the acceleration wasn’t
constant, then we might have something like this instead of the pink straight
line. And in that case, our area
calculations don’t work anymore, because in this particular case, we haven’t
accounted for this extra area. And the point is that if this line
wasn’t a straight line or if we weren’t dealing with a constant acceleration, then
we couldn’t calculate this area using the rectangle plus the triangle area
method. So now that we’ve seen all of this,
let’s take a look at an example question.

A bicycle has an initial velocity
of 12 metres per second. And it accelerates for 15 seconds
in the direction of its velocity, moving 220 metres during that time. What is the bicycle’s rate of
acceleration? Answer to two decimal places.

Okay, so in this question, we’ve
got a bicycle. So let’s say that this is our
bicycle. And we’ve been told that it has an
initial velocity of 12 metres per second. So let’s arbitrarily choose that,
initially, the bicycle is moving towards the right at 12 metres per second. Now let’s also say that its initial
velocity, for simplicity, we will call 𝑢. And let’s write down on the left
that 𝑢 is equal to 12 metres per second. Then we’ve been told that the
bicycle accelerates in the direction of its velocity. In other words, the acceleration is
in the same direction as 𝑢. So let’s say that the acceleration
of the bike is 𝑎. Now this is what we’re trying to
calculate here.

Additionally, we’ve also been told
that the bicycle accelerates for 15 seconds. So let’s say that the bicycle is
here at the beginning of its motion. And then it ends up here. And the time taken for the bicycle
to actually go there we will call 𝑡. We also know that 𝑡 is equal to 15
seconds. And as well as this, we’ve been
told the distance moved by the bike.

Now this distance is the distance
between when it starts accelerating and it finishes accelerating. We’ll call this distance 𝑠. And we know that distance 𝑠 is 220
metres. We’ve been told that in the
question as well. And so, at this point, we know the
value of 𝑢, 𝑡, and 𝑠. And we need to find out what 𝑎 is,
the acceleration of the bike.

To calculate this, we need to
recall one of the kinematic equations. Specifically, the equation that
we’re looking for is this one here. The equation tells us that when an
object moves in a straight line with a constant acceleration, then the displacement
of the object or the shortest distance, the straight-line distance, between its
start and finish points, which in this case is the distance that we’ve been
given. So the equation tells us that that
distance is equal to the initial velocity multiplied by the time over which the
object accelerates. Plus half multiplied by the
acceleration of the object multiplied by the time over which it accelerates
squared.

And so we’re going to use this
equation. And we’re going to rearrange to
solve for the acceleration 𝑎. To do this, we’re first going to
subtract 𝑢𝑡 from both sides of the equation, because this way, on the right-hand
side, the 𝑢𝑡s cancel. And so we’re left with 𝑠 minus
𝑢𝑡 is equal to half 𝑎𝑡 squared. And then we’re going to multiply
both sides of the equation by two, cause this way, on the right, the factor of half
cancels with the two. And so we’re left with two
multiplied by 𝑠 minus 𝑢𝑡 is equal to 𝑎𝑡 squared. Then we divide both sides of the
equation by 𝑡 squared so that, on the right-hand side, it cancels. And so we’re finally left with two
multiplied by 𝑠 minus 𝑢𝑡 divided by 𝑡 squared is equal to the acceleration
𝑎.

Now before we start plugging in any
values to this equation, we can see very quickly that the quantities that we’ve
written down here are all in their base units. Metres per second for velocity,
seconds for time, and metres for displacement. This means that when we plug our
values into our equation, our final answer is going to be in its own base unit. And the base unit of acceleration
is metres per second squared. So at this point, we can plug in
our numbers without having to worry about units.

When we do sub in our numbers, we
get two multiplied by 𝑠 minus 𝑢𝑡 divided by 𝑡 squared. And when we evaluate the fraction,
we find that the acceleration, 𝑎, is equal to 0.35555 recurring metres per second
squared. And when we round our answer to two
decimal places, as we’ve been asked to do in the question, we can say that the
acceleration of the bicycle is 0.36 metres per second squared.

So now that we’ve had a look at an
example question, let’s summarise what we’ve talked about in this lesson. We’ve seen firstly that if an
object accelerates at a constant rate and moves in a straight line, then we can
describe its motion with a kinematic equation. 𝑠 is equal to 𝑢𝑡 plus half 𝑎𝑡
squared. If we happen to forget this
equation, then we can work it out by plotting a velocity versus time graph and
seeing that the area under the graph, which represents the displacement of the
object, is equal to 𝑢𝑡 plus half 𝑎𝑡 squared.

And finally, we should note that
it’s important to label all the directions, whether that be the direction of the
initial velocity, the displacement, the final velocity, or the acceleration. Everything that points in the same
direction should have the same sign. So, for example, we might decide
that anything pointing towards the right is positive. And therefore, anything pointing
towards the left is negative. But once we make this decision, we
need to stick with it. And so that is an overview of how
we can analyse acceleration over distance and time.