 In this video, we provide the solution to question number 12 for practice exam number 3 for Math 12-10. It's a two-part question. We're told that the position of a particle along a straight line is given by the equation s of t is equal to t cubed minus 9t squared plus 24t plus 4, where s is being measured in meters, t is being measured in seconds, and this model is valid from the time stamp 0 seconds up to 10 seconds. So the first question asks us to find the acceleration. So the relationship between position and acceleration is it's the second derivative. Note, of course, that the derivative of position is equal to the velocity function, and the derivative of velocity is equal to acceleration. Acceleration is the change of velocity, which as velocity is the change of position, this gives us the second derivative. So what we then have to do is then compute those things. So we already have the position function, the velocity function, which is equal to the first derivative of position. By the usual derivative rules, we're going to get 3t squared minus 18t plus 24. That's our velocity. Then acceleration, which is the derivative of velocity here, and you can emphasize this as the second derivative of position if you wanted to, we're going to get 6t minus 18. So that is then our acceleration function. Maybe put a box around it to indicate that's the answer to the first part. Now, for the second part here, we're going to, we have to then note over what time intervals is the particle moving forward, okay? We have to write our answer in interval notation, and we're giving a hint here. If the particle is moving forward, what does that mean? If it's moving forward, is the velocity positive or negative? It'll be moving forward when the velocity is positive. If the velocity was negative, that means it's moving backward. Now, moving forward has nothing to do with acceleration. If we're moving forward and we have a negative acceleration, that just means we're slowing down, but we're still moving forward. If we're moving backward, and we have a positive acceleration, that just means we're slowing down, and therefore we're still going backward though, right? Moving forward versus moving back has only to do with the velocity function. If you were asked instead, when is it speeding up or slowing down, then we have to compare velocity and acceleration together. And you have some homework questions that explore that very idea, so I won't go into detail there. So we have to figure out when is the velocity function positive? So we will begin this by solving the equation velocity is equal to zero, which based upon our calculation above, we get 3t squared minus 18t plus 24. When is that equal to zero? Noticing that everything is divisible at 3 on the left-hand side, I'm just going to divide everything by 3. Smaller coefficient is always better. We get t squared minus a 6t plus 8 is equal to zero. This is now a quadratic equation. We could solve it in many various ways. We could try factoring it. We could use the quadratic formula. I think factoring is going to work out just fine here. We need factors of 8 that add up to be negative 6. Since it's positive 8, of course, that means they have to both be negative. We could take 2 and 4. That is to say t minus 2 and t minus 4 as our factors. This tells us that t equals 2 and t equals 4. This is not the answer to the question. This is the moment where we have zero velocity. This will be the moment where velocity equals zero. So these are the moments of time where it is standing still. We really, what we need to solve is not actually when t equals zero. We need to figure out when, excuse me, when v of t equals zero. We need to figure out when v of t is positive. Moving forward coincides with positive velocity. We need to figure out when this happens. There's a couple of ways we could solve this inequality. These two numbers are of critical importance to us. I like to think of it graphically personally because our velocity function, v of t, is a parabola. It would look something like the following. Notice that the leading coefficient is a positive 1 or in this case a positive 3. That means it's going to be a concave up parabola which the x intercepts are going to be exactly these numbers we found here. That's what we're doing it. We have these markers t equals 2 and t equals 4. So where is it above the x-axis? The y-coordinate being positive means you're above the x-axis. So that's going to be happening over here and this is going to be happening over here. So we need to find values less than 2 or greater than 4. So then recording our answer, we're going to get 0 to 2 union 4 to 10. So why do we get this? We want things less than 2 but our model is only valid between 0 and 10 seconds. So we wouldn't go from negative infinity to 2. We'd only go from 0 to 2. So it's moving forward then. And then it's also moving forward from 4 to 10. 10 and 0 are included. We get the brackets there. 2 and 4 are not included because at the moment 2 seconds and 4 seconds the velocity is 0. So it's neither moving forward nor is it moving back. So the correct answer would be the interval 0 to 2 union 4 to 10 where 0 and 10 are included and 2 to 4 are excluded.