Determining Time Intervals Of Rightward Particle Motion

Hey guys! In this article, we're diving into a classic calculus problem: analyzing the motion of a particle given its position function. We'll be using derivatives to figure out when the particle is moving to the right (or in the positive direction). It's like being a detective, but instead of solving crimes, we're solving motion mysteries! Our specific scenario involves the position function s(t) = (t^3)/3 - (12t^2)/2 + 36t, where s is measured in feet and t in seconds, over the time interval t = 0 to t = 15 seconds. The big question we're tackling is: During what time intervals is the particle moving to the right? This means we need to find when the particle's velocity is positive. So, buckle up, and let's get started on this mathematical journey! We'll break down each step, making sure it's super clear and easy to follow. Remember, understanding the concepts is key, and by the end of this article, you'll be a pro at solving these types of problems.

Before we jump into the math, let's quickly review the relationship between position, velocity, and acceleration. Think of it like this: position tells you where the particle is, velocity tells you how fast it's moving and in what direction, and acceleration tells you how the velocity is changing. The position function, s(t), gives the particle's location at any time t. The velocity, v(t), is the rate of change of position, which we find by taking the first derivative of the position function, v(t) = s'(t). The velocity tells us not only the speed but also the direction of movement. A positive velocity means the particle is moving to the right (or in the positive direction), while a negative velocity means it's moving to the left (or in the negative direction). Finally, the acceleration, a(t), is the rate of change of velocity, calculated as the second derivative of the position function, a(t) = v'(t) = s''(t). Acceleration tells us if the particle is speeding up or slowing down. In our case, we're primarily interested in the velocity to determine when the particle is moving to the right. So, we'll be focusing on finding the derivative of the position function and analyzing when it's positive. Understanding these fundamental concepts is crucial for solving problems related to motion in calculus. We will use these concepts to dissect the given position function and extract meaningful information about the particle's movement. Let's move forward and apply these concepts to our specific problem.

Okay, so the first thing we need to do is figure out the velocity function. Remember, the velocity function, v(t), is just the first derivative of the position function, s(t). Our position function is given by s(t) = (t^3)/3 - (12t^2)/2 + 36t. To find the derivative, we'll use the power rule, which states that if we have a term at^n, its derivative is nat^(n-1). Applying this rule to each term in our position function, we get:

• The derivative of (t^3)/3 is (3t^2)/3 = t^2
• The derivative of -(12t^2)/2 which simplifies to -6t^2 is -12t
• The derivative of 36t is 36

So, putting it all together, the velocity function v(t) is: v(t) = t^2 - 12t + 36. Now, we have a quadratic equation representing the velocity of the particle at any time t. This is a crucial step because the velocity function will tell us when the particle is moving to the right (positive velocity) and when it's moving to the left (negative velocity). To figure out when the particle is moving to the right, we need to find the intervals where v(t) > 0. This involves solving for when the quadratic expression is greater than zero. Next, we'll look at how to find those intervals by analyzing the roots of the quadratic equation. We're getting closer to solving the mystery of the particle's movement!

Now that we have the velocity function v(t) = t^2 - 12t + 36, our next step is to find the critical points. These are the points where the velocity is either zero or undefined. In this case, since our velocity function is a polynomial, it's always defined, so we just need to find where the velocity is equal to zero. This is super important because these critical points are where the particle might change direction – from moving to the right to moving to the left, or vice versa. To find these points, we set v(t) = 0 and solve for t: t^2 - 12t + 36 = 0. This is a quadratic equation, and we can solve it by factoring. We're looking for two numbers that multiply to 36 and add up to -12. Those numbers are -6 and -6. So, we can factor the quadratic as: (t - 6)(t - 6) = 0 or (t - 6)^2 = 0. This gives us a single solution: t = 6. This means that at t = 6 seconds, the velocity of the particle is zero. This is a critical point, and it's a key moment in the particle's journey. It's like a turning point in a story! Now that we've found our critical point, we need to figure out what's happening to the velocity before and after this point. This will tell us when the particle is moving to the right. In the next section, we'll use this critical point to analyze the intervals where the velocity is positive.

Alright, we've pinpointed our critical point at t = 6 seconds. Now, the crucial part: determining the intervals where the velocity is positive. This will tell us exactly when the particle is moving to the right. Remember, the velocity function is v(t) = t^2 - 12t + 36 = (t - 6)^2. Notice something special about this function? It's a square! This means that v(t) will always be greater than or equal to zero, because squaring any real number always results in a non-negative value. So, v(t) ≥ 0 for all t. However, we know that at t = 6, v(t) = 0. This means the particle momentarily stops at t = 6, but it never actually moves to the left (negative velocity). Now, let's consider the time interval we're interested in, which is t = 0 to t = 15 seconds. Since the velocity is always non-negative, the particle is either moving to the right or momentarily at rest within this interval. Except for the single point t = 6, where the velocity is zero, the particle is moving to the right for all other times in the interval. Therefore, we can say the particle is moving to the right on the intervals [0, 6) and (6, 15]. It's like the particle is always pushing forward, briefly pausing at t = 6, but then continuing its journey in the same direction. This is a cool result because it shows how analyzing the properties of the velocity function, like being a square, can give us a lot of information about the particle's motion. In the next section, we'll summarize our findings and wrap up this motion mystery.

So, guys, we've successfully navigated the world of particle motion! We started with the position function, s(t) = (t^3)/3 - (12t^2)/2 + 36t, and our mission was to find the intervals where the particle is moving to the right within the time frame t = 0 to t = 15 seconds. We discovered that the key to solving this was finding the velocity function, which we did by taking the derivative of the position function: v(t) = t^2 - 12t + 36. Then, we identified the critical point by setting the velocity function equal to zero and solving for t, which gave us t = 6. The real magic happened when we realized that the velocity function could be written as v(t) = (t - 6)^2. This form revealed that the velocity is always non-negative, meaning the particle is either moving to the right or at rest. At t = 6, the velocity is zero, indicating a momentary pause. But for all other times in the interval [0, 15], the particle is moving to the right. Therefore, the particle is moving to the right on the intervals [0, 6) and (6, 15]. This journey through calculus has shown us how powerful derivatives can be in understanding motion. By analyzing the position and velocity functions, we've uncovered the secrets of the particle's movement. Remember, calculus isn't just about formulas and equations; it's about understanding change and motion in the world around us. Keep exploring, keep questioning, and keep solving those mathematical mysteries! You've got this!