### Video Transcript

A particle is moving in a straight
line such that its displacement 𝑆 at time 𝑡 is given by 𝑆 is equal to one-third
𝑡 cubed minus a quarter 𝑡 squared meters. What is the total distance covered
by the particle during the first two seconds? Write your answer approximate to
two decimal places.

In this question, we are given an
expression for the displacement of a particle 𝑆 in terms of its time 𝑡. We are asked to calculate the total
distance covered during the first two seconds. As a result, we may initially be
tempted to substitute 𝑡 equals two and 𝑡 equals zero into our expression and
subtract the answers. However, this will not take into
account the direction that the particle is moving within the two seconds. It will only calculate the
displacement and not the total distance covered.

We will therefore begin by finding
the times where the velocity of the particle 𝑉 is greater than and less than
zero. We can find an expression for the
velocity by differentiating our expression for the displacement with respect to
𝑡. 𝑉 is equal to d𝑠 by d𝑡. Using the power rule of
differentiation, d𝑠 by d𝑡 is equal to three over three 𝑡 squared minus two over
four 𝑡. And this simplifies to 𝑡 squared
minus one-half 𝑡.

Setting our expression equal to
zero, we can solve the equation by first factoring the quadratic. We have 𝑡 multiplied by 𝑡 minus
one-half is equal to zero. This gives us two values of 𝑡,
where the velocity is equal to zero, 𝑡 is equal to zero, and 𝑡 is equal to
one-half. Substituting any value of 𝑡
between zero and one-half into our expression for 𝑉 gives us a negative answer. And we can therefore conclude that
the velocity is less than zero when 𝑡 is greater than zero and less than
one-half. In a similar way, when 𝑡 is
greater than one-half and less than two, the velocity 𝑉 is greater than zero. We therefore need to find the value
of 𝑆 between 𝑡 equals zero and 𝑡 equals one-half and separately between 𝑡 equals
one-half and 𝑡 equals two.

Subtracting 𝑆 of zero from 𝑆 of
one-half will give us the displacement in the first half a second. And taking the magnitude or
absolute value of this will give us the distance covered in this time. Clearing some space, when 𝑡 is
equal to one-half, 𝑆 is equal to one-third multiplied by a half cubed minus
one-quarter multiplied by a half squared. This is equal to negative one over
48. When 𝑡 is equal to zero, 𝑆 is
equal to zero. This means that the distance
covered between 𝑡 equals zero and 𝑡 equals one-half is the absolute value of
negative one forty-eighth minus zero, which in turn is equal to one
forty-eighths.

We repeat this process for the time
period between 𝑡 equals one-half and 𝑡 is equal to two. When 𝑡 is equal to two, 𝑆 is
equal to one-third multiplied by two cubed minus one-quarter multiplied by two
squared. This simplifies to eight-thirds
minus one, which is equal to five-thirds. The distance traveled between 𝑡
equals one-half and 𝑡 equals two is therefore equal to the absolute value of
five-thirds minus negative one forty-eighth. And this is equal to twenty-seven
sixteenths.

The total distance covered by the
particle during the first two seconds is therefore equal to one forty-eighth plus
twenty-seven sixteenths. And this is equal to forty-one
twenty-fourths. As we are asked to approximate the
answer to two decimal places, the total distance covered during the first two
seconds is 1.71 meters.