### Video Transcript

In this video, we will learn how to
use differentiation to find the instantaneous velocity, speed, and acceleration of a
particle. We will begin by recalling some key
terms, notation, and their units.

The displacement of a body, denoted
π of π‘, is measured in meters. The distance a body travels is the
magnitude of π of π‘, as the distance must always be positive. This also has a standard unit of
meters. The velocity of a body is denoted
π£ of π‘ and is measured in meters per second. As with distance, the speed of a
body must be positive. Therefore, the speed is equal to
the magnitude of the velocity. Finally, we have the acceleration
π of π‘ which has a standard unit of meters per second squared.

Before we consider how we can use
differentiation to solve problems, we will begin by considering a velocityβtime
graph and how this links these terms. The sketch shows a velocityβtime
graph with six distinct periods. We know that the derivative of the
velocity gives us the acceleration of a body. So when the graph slopes upwards,
we have a positive acceleration. And we can calculate this
acceleration by working out the slope or gradient of the graph. Likewise, when the graph slopes
downwards, there is a negative acceleration. Finally, when the graph is
horizontal, there is no slope or gradient. This means that there is zero
acceleration and the body is traveling with constant velocity.

Displacement is the change in
position of an object. And we can calculate this
displacement by working out the area between the graph and the π‘-axis. When the graph is above this axis,
the displacement will be positive, and when it is below the axis, it will be
negative. We will now look at a specific
question involving a velocityβtime graph.

The figure shows a velocityβtime
graph for a particle moving in a straight line. When is the acceleration of the
particle negative?

We recall that when dealing with a
velocityβtime graph, the acceleration is negative when the slope moves downwards
from left to right. This occurs during three periods on
the graph. Between π‘ equals zero and π‘
equals three, the velocity π£ of π‘ decreases from six to zero. This means it is decelerating or
the acceleration is negative. From π‘ equals three to π‘ equals
four, the velocity decreases from zero to negative six. And finally, between π‘ equals nine
and π‘ equals 11, the velocity decreases from five to zero. We can therefore conclude that the
acceleration is negative from π‘ equals zero to π‘ equals four, and it is also
negative from π‘ equals nine to π‘ equals 11. There are two continuous periods on
the graph where the acceleration of the particle is negative.

We will now move on to consider how
we can convert between displacement, velocity, and acceleration using
differentiation. We recall that the displacement of
a body at any given time π‘ can be written as π of π‘. The velocity is π£ of π‘ and the
acceleration π of π‘. If we are given an expression for
π of π‘, we can differentiate with respect to π‘ to find an expression for π£ of
π‘. The same is true to move from the
velocity to acceleration. This means that π£ of π‘ is equal
to d by dπ‘ of π of π‘. Likewise, π of π‘ is equal to d by
dπ‘ of π£ of π‘. We can also find an expression for
the acceleration π of π‘ by differentiating the displacement expression twice.

Whilst it will not be required in
this video, we can move in the opposite direction from π of π‘ to π£ of π‘ and from
π£ of π‘ to π of π‘ by integrating. This is because integration is the
opposite or inverse of differentiation. We will now look at some examples
where we need to find the expression for the velocity or acceleration of a body and
extend this to find the velocity or acceleration at a certain time.

A particle is moving in a straight
line such that its displacement π after π‘ seconds is given by π is equal to two
π‘ squared minus three π‘ plus three meters, where π‘ is greater than zero. Determine the velocity π£ as a
function of time.

We are told in the question that
the displacement of a particle π is equal to two π‘ squared minus three π‘ plus
three meters. And we need to find an expression
for the velocity π£. In order to do this, we will need
to differentiate our function, as π£ of π‘ is equal to d by dπ‘ of π of π‘. If the displacement of a body is
given as a function in terms of time, we can differentiate to find an expression for
the velocity. Differentiating two π‘ squared
gives us 4π‘. Differentiating negative three π‘
gives us negative three. And differentiating a constant, in
this case three, gives us zero. The velocity is therefore equal to
4π‘ minus three. As we are differentiating with
respect to π‘, π£ is equal to 4π‘ minus three meters per second. This is an expression for the
velocity as a function of time.

In our next question, we need to
find an expression for the acceleration as a function of time.

A particle is moving in a straight
line such that its displacement π after π‘ seconds is given by π is equal to two
π‘ cubed plus two π‘ plus two meters, where π‘ is greater than zero. Determine the acceleration π as a
function of time.

In order to answer this question,
we recall that in order to find an expression for the velocity, we can differentiate
an expression in terms of time for the displacement. Likewise, we can differentiate the
velocity expression to calculate an expression for the acceleration. This means that π£ of π‘ is equal
to d by dπ‘ of π of π‘. Likewise, π of π‘ is equal to d by
dπ‘ of π£ of π‘. In this question, we are told that
the displacement π is equal to two π‘ cubed plus two π‘ plus two meters.

Differentiating this with respect
to π‘ will give us an expression for the velocity. Differentiating two π‘ cubed gives
us six π‘ squared. Differentiating two π‘ gives us
two, and differentiating the constant two gives us zero. Therefore, π£ is equal to six π‘
squared plus two. As the displacement is measured in
meters and the time in seconds, the velocity will be measured in meters per
second. π£ is equal to six π‘ squared plus
two meters per second.

Differentiating this new expression
with respect to π‘ will give us an expression for π. When we differentiate six π‘
squared, we get 12π‘, and once again differentiating a constant gives us zero. The acceleration π is therefore
equal to 12π‘. We can therefore conclude that if
π is equal to two π‘ cubed plus two π‘ plus two meters, then the acceleration π of
the particle as a function of time is equal to 12π‘ meters per second squared.

In our next question, we need to
calculate the magnitude of the acceleration of a particle when it is traveling at a
particular velocity.

A particle is moving in a straight
line such that its velocity π£ at time π‘ seconds is given by π£ is equal to two π‘
squared minus 68 meters per second, where π‘ is greater than or equal to zero. Find the magnitude of the
acceleration of the particle when its velocity is 94 meters per second.

To answer this question, we recall
that in order to find an expression for the acceleration of a particle π of π‘, we
differentiate the expression π£ of π‘. The velocity of this particle at
time π‘ seconds is equal to two π‘ squared minus 68 meters per second. Differentiating two π‘ squared
gives us four π‘. Differentiating any constant gives
us zero. Therefore, differentiating negative
68 is zero. The acceleration π is therefore
equal to four π‘. As the time π‘ is measured in
seconds, the acceleration is equal to four π‘ meters per second squared.

We are asked to calculate the
magnitude of this acceleration when the velocity is 94 meters per second. We can do this by finding the value
of π‘ for which the velocity is 94 meters per second and then substitute this value
of π‘ into our equation for π of π‘. Setting our equation for π£ equal
to 94, we have two π‘ squared minus 68 equals 94. Adding 68 to both sides of this
equation, we have two π‘ squared is equal to 162. We can then divide both sides of
the equation by two so that π‘ squared is equal to 81. Square rooting both sides of this
equation, we have π‘ is equal to the positive or negative square root of 81. As nine squared equals 81, this
means that π‘ is equal to positive or negative nine.

We are told in the question as π‘
stands for time that π‘ is greater than or equal to zero. This means that the time at which
the velocity is 94 meters per second is nine seconds. We can now substitute this value
into our expression for the acceleration. π is equal to four multiplied by
nine. As four multiplied by nine is 36,
we can conclude that the magnitude of the acceleration of the particle when the
velocity is 94 meters per second is 36 meters per second squared.

We will now summarize the key
points from this video. We saw in this video that we can
use differentiation to find the velocity and acceleration of a particle as a
function of time. If we have an expression for the
displacement of the particle or body in terms of time, denoted π of π‘, we can find
an expression for π£ of π‘, the velocity of the particle, by differentiating. Likewise, we can differentiate this
expression π£ of π‘ with respect to time to find an expression for π of π‘, the
acceleration. This can be written more formally
as shown.

The speed of a particle is the
magnitude of its velocity, and the distance is the magnitude of the
displacement. This is because both speed and
distance must be positive values. Whilst we didnβt cover it in this
video, it is also important to note that a particle is instantaneously at rest when
π£ of π‘ is equal to zero.