### Video Transcript

In this video, we’ll learn how to
apply the laws of the uniform acceleration motion of a particle in a straight
line. Up until now, we’ve used the
speed–distance–time formula. Speed is distance divided by
time. That formula is applicable when the
acceleration of the object is equal to zero. We might also have considered
acceleration in terms of the gradient or the slope of the line on a velocity–time
graph. Now, we look to introduce some new
equations. These equations are sometimes
called the equations of constant or uniform acceleration, so called because they’re
applied when working with uniform acceleration, that is, acceleration that doesn’t
change. So, let’s see where these equations
come from.

We begin by considering a
velocity–time graph. 𝑣 naught or 𝑣 sub zero is the
initial velocity or the original velocity of our object. 𝑣, then, is the velocity after a
time 𝑡. We let 𝑎 be equal to
acceleration. And Δ𝑥 or change in 𝑥 is the
displacement of the object, that sometimes given as 𝑠. Now, we know that the acceleration
is given by the slope of the line on a velocity–time graph. Since the acceleration is constant,
we have a straight line. And to find the slope or gradient
of a straight line, we use the formula rise over run or change in 𝑦 over change in
𝑥.

In our diagram, change in 𝑦 is
this length here. It’s the difference between the
initial velocity and the velocity after 𝑡 time units. That’s 𝑣 minus 𝑣 naught. Change in 𝑥 is the length of this
line. That’s 𝑡 minus zero or just
𝑡. And so, the slope of our line which
represents the acceleration 𝑎, under these conditions, is given by 𝑣 minus 𝑣
naught over 𝑡. If we multiply both sides of the
equation by 𝑡, we get 𝑎𝑡 equals 𝑣 minus 𝑣 naught. And then if we add 𝑣 naught to
both sides, we get 𝑣 equals 𝑣 naught plus 𝑎𝑡. So 𝑣 equals 𝑣 naught plus
𝑎𝑡. That’s our first kinematic
equation.

We’ll now consider our second. This time we go back to our graph
and recall that the displacement is found by calculating the area between the line
and the 𝑥-axis. We can split this area into a
rectangle and a triangle. The area of the rectangle is the
product of its dimensions. So that’s 𝑣 naught times 𝑡. And the area of a triangle is a
half times the base times the height. So here, that’s a half times 𝑣
minus 𝑣 naught times 𝑡. And so, this means that the
displacement Δ𝑥, which is of course the area between the line and the 𝑥-axis, is
𝑣 naught 𝑡 plus a half times 𝑣 minus 𝑣 naught 𝑡. But earlier, we defined 𝑣 minus 𝑣
naught as being equal to 𝑎𝑡. So, we replace this with 𝑎𝑡 in
our equation, and we get Δ𝑥 equals 𝑣 naught 𝑡 plus a half times 𝑎𝑡 times 𝑡 or
a half 𝑎𝑡 squared. And this is our second kinematic
equation.

Now, it’s outside the scope of this
video to derive all of our equations. But by either rearranging the two
we have or considering another graphical approach, we’re able to obtain two further
formulae. Those are Δ𝑥 equals a half times
𝑣 naught plus 𝑣 times 𝑡 and 𝑣 squared equals 𝑣 naught squared plus two times 𝑎
times Δ𝑥. So, how do we apply these
formulae? We begin by listing all of the
measurements that we know. We then identify the one that we
don’t know and are not trying to find. That will eliminate all of the
equations we aren’t interested in, leaving just one behind. Let’s see what that looks like.

If a particle started moving in a
straight line with an initial velocity of 25.1 centimeters per second and a uniform
acceleration of 2.4 centimeters per square second, determine its velocity after nine
seconds.

The question states that the
particle has uniform acceleration. This is a good indication to us
that we’re going to need to use the equations of uniform or constant
acceleration. Those are the four kinematic
equations. So, we begin by listing the
equations out. Our job is to eliminate all but one
of these equations. And we do so by listing the
measurements that we’ve been given in the question. We’re given the initial velocity 𝑣
naught. That’s 25.1 centimeters per
second. We’ve got an acceleration of 2.4
centimeters per square second and a time nine seconds.

We’re looking to calculate the
velocity after nine seconds. Notice that that means we’re really
not interested in Δ𝑥, the displacement of our object. And so, we go through and eliminate
all of the equations that contain Δ𝑥. Those are two, equation three, and
four. And so, we’re left with one
equation; that’s 𝑣 equals 𝑣 naught plus 𝑎𝑡. Next, we substitute everything we
know about our particle into this formula. We’re looking to calculate 𝑣, so
we say that 𝑣 is equal to 𝑣 naught, which is 25.1, plus 𝑎 times 𝑡, that’s 2.4
times nine. 2.4 multiplied by nine is 21.6. So, velocity is given by 25.1 plus
21.6, which is 46.7.

Notice that we’ve been working in
centimeters per second, centimeters per square second, and seconds. And so, the units for our velocity
after nine seconds are centimeters per second. The velocity of the particle is
therefore 46.7 centimeters per second.

In our next example, we’ll consider
how to calculate the distance covered by an object.

A small ball started moving
horizontally at 16.3 meters per second. It moved in a straight line with a
uniform deceleration of three meters per square second. Determine the distance the ball
covered in the first two seconds.

The ball has a uniform
deceleration. In other words, it has a uniform
negative acceleration. That’s an indication to us that
we’re going to use the kinematic equations, the equations of uniform
acceleration. And so, the first thing that we do
is list those four equations. For an initial velocity 𝑣 naught,
an acceleration of 𝑎, a velocity of 𝑣 at time 𝑡 time units, and a displacement of
Δ𝑥, they are as shown.

Our next step is to eliminate all
but one of these equations, and we do so by listing everything we know about the
motion of our object. We’re told that the small ball
starts moving at 16.3 meters per second, so that’s its initial velocity. It has a uniform deceleration of
three meters per square second. In other words, it’s slowing
down. So, we say its acceleration is
negative three. Time 𝑡 is two seconds, and we’re
looking to find the distance that the ball covers.

We recall that distance is simply
the magnitude of the displacement. So, distance here is Δ𝑥. That’s change in 𝑥. And that means we’re not interested
in 𝑣, the final velocity of the object. So, we go through our equations and
we eliminate every equation that contains 𝑣. That’s one, three, and four,
leaving us, of course, with just equation two.

Our next job is to substitute what
we know about our ball into this equation. Δ𝑥 is what we’re trying to
calculate. Then, 𝑣 naught 𝑡 is 16.3 times
two. A half 𝑎𝑡 squared becomes a half
times negative three times two squared. Then, 16.3 times two is 32.6. And a half times negative three
times two squared is negative six. So, Δ𝑥 is 32.6 minus six, which is
26.6, and we’re measuring that in meters. So, the ball covers a distance of
26.6 meters in the first two seconds of its motion.

Now, it might seem that the
information about the ball moving in a straight line is somewhat superfluous. However, it’s actually really
important. It allows us simply to consider one
direction of motion. If we had to consider two
directions, that would complicate things a little more.

Let’s move on to another
example.

A particle moving in a straight
line was accelerating at a rate of 22 centimeters per square second in the same
direction as its initial velocity. If the magnitude of its
displacement 10 seconds after it started moving was 29 meters, calculate the
magnitude of its initial velocity 𝑣 naught and its velocity 𝑣 at the end of this
period.

We’re given that the particle is
accelerating at a constant rate of 22 centimeters per square second. This means to answer this question,
we’re going to need to use the kinematic equations. These are, of course, equations of
constant acceleration. For a starting velocity 𝑣 naught,
a velocity 𝑣 after 𝑡 time units, an acceleration 𝑎, and a displacement Δ𝑥, they
are as shown.

What we do is begin by listing
everything we know about our motion. We’ve already said we know that the
acceleration is constant, and it’s 22 centimeters per square second. It’s in the same direction as its
initial velocity. Now, we don’t know its initial
velocity, but by assuming that they’re in the same direction, we can take both
acceleration and 𝑣 naught to be positive. We’re also told that the magnitude
of the displacement 10 seconds after it started moving was 29 meters. Remember, displacement can have a
direction. So, by considering just the
magnitude, we’re thinking about the distance; that’s 29 meters. Time 𝑡 is 10 seconds.

Now, the question actually asks us
to calculate the magnitude of the initial velocity and its velocity at the end of
the period. Let’s begin by calculating its
initial velocity 𝑣 naught. In this case, we’re not interested
in 𝑣, so we go through our equations and eliminate those containing 𝑣. Those are one, three, and four. Our next step would normally be to
substitute everything we know about the motion of our particle into that second
equation. We do have a little bit of a
problem though. We notice that the units for our
acceleration and our displacement are different. We need them to be the same. So, we multiply displacement by 100
and we find it’s actually equal to 2900 centimeters.

Then, substituting everything we
know into this formula, and we get 2900 equals 10𝑣 naught plus a half times 22
times 10 squared. A half times 22 times 10 squared is
1100. So, we subtract 1100 from both
sides, and we find that 1800 is equal to 10 times 𝑣 naught. Our final step is to divide through
by 10. 1800 divided by 10 is 180. Now, we’re working in
centimeters. So, our velocity, our initial
velocity 𝑣 naught, is 180 centimeters per second. We might choose to give our answer
in meters per second by dividing through by 100. And when we do, we find that 𝑣
naught is 1.8 meters per second.

We’re not quite finished. We’re still looking to calculate
its velocity 𝑣 at the end of the period. Now that we know 𝑣 naught, we can
actually use any of our equations. So, let’s use the first one. We substitute everything we know
about the motion of our particle into this formula, continuing to work in
centimeters and centimeters per second. When we do, we get 𝑣 is 180 plus
22 times 10. 22 times 10 is 220. And 180 plus 220 is 400. We’re still, of course, working in
centimeters per second. To give our answer in meters per
second, we’ll divide through by 100. And when we do, we find that the
velocity 𝑣 at the end of the motion is four meters per second.

In our final example, we’ll look at
finding the final velocity of the body.

A particle was moving in a straight
line with a constant acceleration of two centimeters per square second. Given that its initial velocity was
60 centimeters per second, find the velocity of the body when it was 15 meters from
the starting point.

We have a constant acceleration, so
we’re going to use our kinematic equations. For a starting velocity 𝑣 naught,
an acceleration 𝑎, and a velocity 𝑣 after 𝑡 time units, the first equation is 𝑣
equals 𝑣 naught plus 𝑎𝑡. We introduce Δ𝑥 to represent the
displacement of the objects. And we get Δ𝑥 equals 𝑣 naught 𝑡
plus a half 𝑎𝑡 squared. Our third equation Δ𝑥 is equal to
a half 𝑣 naught plus 𝑣 times 𝑡. And then we have one further
equation. That is 𝑣 squared equals 𝑣 naught
squared plus two times 𝑎 times Δ𝑥.

Let’s list what we know about the
motion of our particle. We have acceleration as two
centimeters per square second, an initial velocity of 60 centimeters per second, and
a displacement Δ𝑥 of 15 meters. Now, in fact, our acceleration and
velocity are given in centimeters per second and centimeters per square second. So, we want the units to be the
same for Δ𝑥. We multiply 15 by 100 and we see
that that’s 1500 centimeters. We’re looking to find the velocity
of the body when it’s this distance away from the starting point. Notice that that means we’re not
interested in the time that this takes.

So, we go through to our four
kinematic equations and we get rid of those that contain 𝑡. Well, that’s equation one, two, and
three. That leaves us with just one
equation, 𝑣 squared equals 𝑣 naught squared plus two times 𝑎 times Δ𝑥. We’re going to substitute
everything we know about the motion of our particle into this equation. When we do, we get 𝑣 squared
equals 60 squared plus two times two times 1500. 60 squared is 3600. And two times two times 1500 is
6000. So, 𝑣 squared is equal to
9600.

It follows that we can solve for 𝑣
by finding the square root of both sides of our equation. 𝑣 is therefore equal to the square
root of 9600. That’s 97.97 and so on. Correct to the nearest whole
number, that’s 98. And we can therefore say that the
velocity of the body when it’s 15 meters from the starting point is 98 centimeters
per second.

In this video, we’ve learned that
we can use kinematic equations to model motion involving constant acceleration in a
straight line. The four equations we use are
shown. In these equations, 𝑣 naught is
the initial velocity, 𝑣 is the velocity after 𝑡 time units, 𝑎 is the constant
acceleration, and Δ𝑥 is the displacement of the object. We also saw that it’s really
important when working with these equations to make sure all the units are the
same.