### Video Transcript

A particle moves in a straight line
such that at time 𝑡 seconds, where 𝑡 is greater than or equal to zero, its
displacement from a fixed point on the line is given by 𝑠 equals negative two 𝑡
squared minus five 𝑡 plus 12 meters. Determine the particle’s velocity
when its displacement is equal to zero.

Here, we’ve been given an
expression for 𝑠 in terms of time 𝑡. We’re asked to find the velocity of
the particle when its displacement is zero. And so to do that, we’re going to
need to find the time 𝑡 at which the displacement is zero. Displacement in this question is
given by 𝑠. So we’re going to let 𝑠 be equal
to zero. When we do, our equation becomes
zero equals negative two 𝑡 squared minus five 𝑡 plus 12.

Now, this equation will be much
easier to solve if the coefficient of 𝑡 squared is positive. And so we’re going to begin simply
by multiplying through by negative one. This essentially has the effect of
changing all of the signs. So we get an equation zero equals
two 𝑡 squared plus five 𝑡 minus 12.

We could use an equation solver on
a calculator. But if we’re going to hand solve
this, we need to begin by factoring the expression two 𝑡 squared plus five 𝑡 minus
12. It’s a quadratic equation. And the three terms are coprime;
they share no other factors than one. And so this means it factors into a
pair of parentheses. The first terms are two 𝑡 and
𝑡. Then we need to consider factors of
negative 12, bearing in mind that one of these is going to be multiplied by two
before we add them to give us a value of five.

And so to achieve this result, our
two binomials need to be two 𝑡 minus three and 𝑡 plus four. And we know that the product of
these two binomials is zero. So what does this tell us? Well, for this statement to be
true, for the product of the binomials to be zero, either one or other of the
binomials must itself be zero. So either two 𝑡 minus three is
zero or 𝑡 plus four is zero. We solve the first equation for 𝑡
by first adding three to both sides, giving us two 𝑡 equals three. And then we divide through by two,
giving us 𝑡 equals three over two or 1.5 seconds.

To solve our second equation, we
subtract four from both sides. And that gives us 𝑡 is equal to
negative four. Well, we can actually disregard
this result, and that’s because we’re told that 𝑡 must be greater than or equal to
zero. And in terms of a context, it
really doesn’t make sense to have a negative time. So we know that the displacement is
zero when 𝑡 is equal to three over two or 1.5 seconds.

But how do we use this to find the
particle’s velocity? Well, we begin by defining velocity
as rate of change of displacement. And when we think about rate of
change, we think about the derivative. In fact, we can say that velocity
𝑣 is the derivative of 𝑠 with respect to time 𝑡. So we’re going to differentiate our
expression for 𝑠. This consists of three terms. So we’ll simply do it term by
term.

The first term we need to
differentiate is negative two 𝑡 squared. And we know that to differentiate a
power term, we multiply the entire term by the exponent and then reduce that
exponent by one. So the first term negative two 𝑡
squared differentiates to two times negative two 𝑡, which is negative four 𝑡. Then, when we differentiate
negative five 𝑡, we simply get negative five. And the derivative of a constant is
zero. So we found an expression for
𝑣. It’s negative four 𝑡 minus
five.

It’s worth noting that since we’re
working in time 𝑡 seconds and the displacement is in meters, velocity will be in
meters per second. We want to calculate the velocity
when its displacement is zero. And we said that happens when 𝑡 is
equal to three over two seconds. So we’ll substitute three over two
into our expression for 𝑣. That’s negative four times three
over two minus five. We can cross cancel by dividing
through by two, and we get negative two times three over one minus five, which is
negative six minus five, or negative 11.

And so we see that when the
particle’s displacement is zero, its velocity is negative 11 meters per second.