# Question Video: Determining the Velocity of a Particle Based on Time and Displacement Mathematics

A particle moves in a straight line such that at time 𝑡 seconds (𝑡 ≥ 0) its displacement from a fixed point on the line is given by 𝑠 = (−2𝑡² − 5𝑡 + 12) m. Determine the particle’s velocity when its displacement is 0.

04:19

### 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.