### Video Transcript

A particle moves along the π₯-axis. At time π‘ seconds, its displacement from the origin is given by π₯ equals ππ‘ squared minus π‘ plus π meters, for π‘ is greater than or equal to zero. When π‘ is equal to one second, π₯ is equal to seven meters, and when π‘ is equal to two seconds, the particleβs velocity is seven meters per second. Determine the value of π minus π.

In order to be able to calculate the value of π minus π, it should be quite clear to us that weβre going to begin by working out what π and π are worth. And so, letβs use the information about the displacement from the origin. At time π‘, the displacement π₯ is ππ‘ squared minus π‘ plus π meters. We also know that when π‘ is equal to one second, π₯ is equal to seven. So, letβs simply begin by substituting π‘ equals one and π₯ equals seven into our equation for displacement. When we do, we get seven equals π times one squared minus one plus π. That right-hand side simplifies to π minus one plus π. And then if we add one to both sides, we have π plus π being equal to eight.

So, how does this help us? Well, we have further information. Weβre told that when π‘ is equal to two seconds, the velocity of the particle, letβs call that π£, is seven meters per second. And so, presumably, if we can find an expression for π£, velocity, and substitute these values in, weβll get a second equation for π and π, which we can then use to solve simultaneously. But how do we find an expression for velocity? Well, velocity is defined as the rate of change of position or displacement. When we think about rate of change, we should be thinking about differentiating. And so, we can say that velocity is the first derivative of displacement with respect to time π‘.

So, letβs differentiate our expression for π₯. We know π and π are constants, so weβll treat them as such. Letβs do this term by term. We begin by differentiating ππ‘ squared. Now, we know that when we differentiate a power term, we multiply the entire term by the exponent and then reduce that exponent by one. So the derivative of ππ‘ squared is two times ππ‘ to the power of one or two times ππ‘. Now, we can either add the derivative of negative π‘ or subtract the derivative of π‘. That constant, negative one, makes no difference otherwise. So, letβs subtract the derivative of π‘.

When we differentiate π‘ with respect to π‘, we simply get one. So, we have an equation for π£; itβs π£ equals two ππ‘ minus one. And weβre now ready to substitute the values of π‘ equals two and π£ equals seven into this equation. When we do, we get seven equals two times π times two minus one. This right-hand side then simplifies to four π minus one. This is simply an equation we can solve in the usual way. We begin by adding one to both sides, and the equation becomes eight equals four π. And then we divide through by four. Eight divided by four is two, so we find π is equal to two.

And weβre now ready to substitute that into our first equation, eight equals π plus π. When we do, our equation becomes eight equals two plus π. To solve for π, we subtract two from both sides, and we get π equals eight minus two, which is equal to six. Now, the question wants us to determine the value of π minus π. Well, we just calculated π to be six and π to be equal to two. So we see that π minus π is six minus two, which is simply four. Given the conditions regarding the motion of our particle, we find π minus π must be equal to four.