### Video Transcript

A particle is moving along the
π₯-axis. For π‘ is greater than or equal to
zero and less than or equal to eight, the acceleration of the particle is given by
π of π‘ equals two π‘ cos of natural log of π‘ squared plus one over π‘ squared
plus one, where π‘ is measured in seconds and π of π‘ in meters per square
second. At time π‘ equals zero, the
velocity of the particle is given by π£ equals negative 2.1 meters per second. i)
Find the velocity of the particle at π‘ equals three. ii) For π‘ is greater than or
equal to zero and less than or equal to eight seconds, at what time interval is the
speed of the particle increasing? iii) For zero is less than or equal to π‘ which is
less than or equal to eight seconds, what is the average velocity of the
particle? And iv) For zero is less than or
equal to π‘ which is less than or equal to eight seconds, find the total distance
travelled by the particle.

In this question, weβve been given
a function for acceleration in terms of time. The first part of this question
asks us to find the velocity of the particle when π‘ is equal to three. So we begin by recalling the
relationship between displacement, velocity, and acceleration. Velocity is the rate of change of
displacement with respect to time. So if π is a function in π‘ where
π is displacement, then π£ of π‘ is equal to π prime of π‘. We differentiate our function for
displacement with respect to time.

Similarly, acceleration is the rate
of change of velocity with respect to time. So itβs π£ prime of π‘. Itβs the derivative of π£ with
respect to π‘. We can reverse this process and we
see that we can find a function for the velocity by integrating the function for
acceleration with respect to time. Similarly, we can find a function
for displacement by integrating a function for velocity. To find the velocity when π‘ is
equal to three, weβre going to need to integrate our function π with respect to π‘
and then evaluate it when π‘ is equal to three. So letβs integrate two π‘ cos of
natural log of π‘ squared plus one over π‘ squared plus one.

Weβre going to use integration by
substitution to complete this step. And weβre going to let π’ be equal
to ln π‘ squared plus one. π’ is a composite function. Itβs a function of a function. So weβre going to use the chain
rule to find dπ’ by dπ‘. This time, if we let π£ be equal to
π‘ squared plus one, then π’ is equal to the natural log of π£. And dπ’ by dπ‘ is equal to dπ£ by
dπ‘ multiplied by dπ’ by dπ£. dπ£ by dπ‘ is equal to two π‘ and dπ’ by dπ£ is equal
to one over π£. So dπ’ by dπ‘ is equal to two π‘
times one over π£.

And of course, we said that π£ was
equal to π‘ squared plus one. So dπ’ by dπ‘ is equal to two π‘
times one over π‘ squared plus one. We simplify this a little and say
that dπ’ by dπ‘ equals two π‘ over π‘ squared plus one. And then, we rearrange this a
little. And we see that dπ’ is equal to two
π‘ over π‘ squared plus one dπ‘.

We can therefore replace two π‘
over π‘ squared plus one dπ‘ in our integral with dπ’. And we said that π’ was equal to
natural log of π‘ squared plus one. So weβre going to need to integrate
cosine of π’ with respect to π’. Well, the integral of cosine of π’
is sin π’. And finally, we replace π’ with the
natural log of π‘ squared plus one. And we see that our function for
velocity at time π‘ seconds is sin of the natural log of π‘ squared plus one plus π
the constant of integration.

We cannot evaluate the velocity of
our particle at π‘ equals three without working out what that constant of
integration is. So we use the fact that when π‘ is
equal to zero π£ is equal to negative 2.1. And we get that negative 2.1 equals
sin of the natural log of zero squared plus one plus π. Well, the natural log of zero
squared plus one is zero and sin of zero is also zero. So we see that π is equal to
negative 2.1. And our function for velocity at
time π‘ seconds is sin of the natural log of π‘ squared plus one minus 2.1.

We now have enough information to
find the velocity of the particle when π‘ is equal to three. Itβs sin of the natural log of
three squared plus one minus 2.1 which is negative 1.356 and so on. Correct to three decimal places the
velocity at π‘ equals three is negative 1.356 meters per second.

Part two asks us to calculate the
time interval during which the speed of the particle is increasing. Now, we have to be careful. Speed is not quite the same as
velocity. Velocity is a vector quantity,
whereas speed is a scalar quantity. Itβs called the magnitude of the
velocity. Now, we know that sin of the
natural log of π‘ squared plus one will always be less than or equal to one and
greater than or equal to negative one. And if we take either of these
extreme values and subtract 2.1, we see that our velocity will always be less than
zero. Itβs always negative.

And whilst at first glance, it may
appear that we need the acceleration of the particle to be greater than zero for the
velocity to be increasing, weβre actually interested in its speed which is the
magnitude of velocity. So the magnitude of our velocity,
the speed, is increasing when the acceleration is less than zero. In other words, two π‘ cos of the
natural log of π‘ squared plus one over π‘ squared plus one must be less than
zero. Well, we know that two π‘ is
greater than zero in the interval π‘ is greater than or equal to zero and less than
or equal to eight, as is π‘ squared plus one.

So we need to work out when cosine
of the natural log of π‘ squared plus one is less than zero. What we know that cosine of π is
negative when π is between π by two and three π by two radians. So this means we need to find out
when the natural log of π‘ squared plus one is greater than π by two and less than
three π by two. We make each part of this
inequality an exponent of π. And this means we see that π‘
squared plus one is greater than π to the π by two and less than π to the three
π by two. We solve this equation by
subtracting one from each side and then finding the square root. And that tells us that π‘ is
greater than 1.952 and less than 10.503 seconds.

Well, we were told that π‘ has to
be greater than or equal to zero and less than or equal to eight. So we can see that the speed which
is the magnitude of our velocity is increasing when π‘ is greater than 1.95 seconds
and less than or equal to eight.

In part three, weβre being asked to
find the average velocity of the particle when π‘ is greater than or equal to zero
and less than or equal to eight. Well, we know that we can find the
average velocity by dividing the total displacement by the time taken. So weβre going to need to work out
what the total displacement of the particle is. We can find a function to describe
the displacement by integrating our function π£ with respect to π‘. In fact, weβre going to evaluate
the integral of sin of natural log of π‘ squared plus one minus 2.1 with respect to
π‘ between the limits zero and eight.

Because weβre evaluating this
integral between the limits of zero and eight, we donβt need to worry about the
constant of integration. And we can type this integral
directly into our calculator to get the value of negative 15.797 and so on. The total displacement is therefore
negative 15.797 and the time it took was eight seconds. So velocity is negative 15.797
divided by eight which is negative 1.9746 and so on. Correct to three decimal places,
thatβs negative 1.975 meters per second.

Part four asks us to find the total
distance travelled by the particle between π‘ is greater than or equal to zero and
less than or equal to eight seconds. And youβll notice I have left some
working from part three out. And thatβs because displacement is
related to distance. In fact, just as speed is magnitude
of velocity, distance is the magnitude of the displacement. Since the displacement of the
particle was negative 15.797 and so on meters, the magnitude of the displacement and
thus the distance travelled was 15.797 meters correct to three decimal places.