Particle Motion: Calculating Acceleration & Distance

Hey guys! Let's dive into a classic physics problem. We're going to break down the motion of a particle, focusing on how to calculate its acceleration and the distance it covers. It's all based on the equation: x = 5t³ + 8t² - 7t + 15, where x is in meters (m) and t is in seconds (s). This equation describes the particle's position over time. So, buckle up, because we are going to explore the world of motion and formulas. This stuff can seem tricky at first, but trust me, once you grasp the basics, it becomes pretty cool. Let's get started with our particle motion analysis, where we'll unravel how to find the acceleration and distance traveled by this tiny object.

Finding Acceleration as a Function of Time

Alright, first things first: we need to figure out the acceleration of the particle as time goes on. Acceleration, in simple terms, is how quickly the particle's velocity is changing. To get there, we need to take a few steps, starting with the particle's position equation. The position equation, x = 5t³ + 8t² - 7t + 15, tells us where the particle is at any given moment. From the position equation, we can find the velocity by taking the first derivative with respect to time (t). Mathematically, this means applying the power rule of differentiation. This rule states that the derivative of t^n is nt^(n-1)*. Doing this for each term in our position equation gives us the velocity equation. Let's break it down:

• Original Equation (Position): x = 5t³ + 8t² - 7t + 15
• First Derivative (Velocity): v = dx/dt = 15t² + 16t - 7 (Applying the power rule to each term)

So, the velocity of the particle is v = 15t² + 16t - 7. But we're not done yet; we still need the acceleration. Acceleration is the rate of change of velocity, which means we need to take the derivative of the velocity equation with respect to time again. Applying the power rule one more time allows us to determine the acceleration equation. Here's how that works:

• Velocity Equation: v = 15t² + 16t - 7
• Second Derivative (Acceleration): a = dv/dt = 30t + 16 (Applying the power rule again)

Therefore, the acceleration of the particle is a = 30t + 16. This equation tells us the acceleration at any time t. So, as time passes, the acceleration changes linearly because the acceleration is dependent on t. We've now successfully determined the acceleration as a function of time. We used derivatives, which are essentially a way of finding the instantaneous rate of change of a function at any given point. Pretty neat, right? Now, let's look at the distance traveled.

Calculating the Distance Traveled in the First 2 Seconds

Now, let's shift gears and calculate how far the particle travels during the first 2 seconds. This isn't as simple as just plugging t = 2 into our position equation, because the particle could change direction during that time. To accurately determine the distance traveled, we need to consider the particle's movement direction. The distance traveled can be different from the displacement, which is the change in the particle's position. Here's a quick guide of how we should approach the problem:

1. Find the Time When Velocity is Zero: The particle changes direction when its velocity momentarily becomes zero. So, we first need to determine the time(s) when the velocity, v = 15t² + 16t - 7, equals zero. This involves solving a quadratic equation.
2. Determine the Position at Critical Times: We will calculate the particle's position at the beginning (t = 0), at the time(s) when the velocity is zero (if any fall within our 0-2 second interval), and at the end of the time interval (t = 2).
3. Calculate the Distance Traveled: Finally, to get the total distance, we add up the absolute values of the changes in position between those critical times. This accounts for any direction changes.

Let's apply these steps. First, we need to find the time when the velocity is zero. We solve the quadratic equation 15t² + 16t - 7 = 0. You can use the quadratic formula to solve for t. The quadratic formula is t = (-b ± √(b² - 4ac)) / 2a. In our velocity equation, a = 15, b = 16, and c = -7. Plugging in those values, we get two possible values for t: t ≈ 0.35 s and t ≈ -1.35 s. Since we're only interested in the time interval from 0 to 2 seconds, we only care about t ≈ 0.35 s. Next, we'll calculate the position of the particle at t = 0 s, t ≈ 0.35 s, and t = 2 s by using the position equation x = 5t³ + 8t² - 7t + 15.

• At t = 0 s: x = 5(0)³ + 8(0)² - 7(0) + 15 = 15 m
• At t ≈ 0.35 s: x ≈ 5(0.35)³ + 8(0.35)² - 7(0.35) + 15 ≈ 13.78 m
• At t = 2 s: x = 5(2)³ + 8(2)² - 7(2) + 15 = 79 m

Now, we can find the distance traveled. From t = 0 to t ≈ 0.35 s, the particle moves from 15 m to about 13.78 m, covering a distance of approximately |13.78 - 15| = 1.22 m. From t ≈ 0.35 s to t = 2 s, the particle moves from 13.78 m to 79 m, covering a distance of approximately |79 - 13.78| = 65.22 m. The total distance traveled in the first 2 seconds is the sum of these distances, which is 1.22 m + 65.22 m ≈ 66.44 m. So, in the first 2 seconds, the particle travels approximately 66.44 meters. Cool, right? It shows how a little bit of math and physics can help us understand the movement of the particle.

Conclusion

Alright, folks, we've successfully navigated the world of particle motion! We determined the acceleration of the particle as a function of time (a = 30t + 16) and calculated the distance it traveled in the first 2 seconds (approximately 66.44 meters). Remember, understanding derivatives is key to tackling these kinds of problems, as they allow us to analyze how things change over time. The concept of zero velocity is also crucial when determining distances, as it helps us identify points where the direction of motion might change. Keep practicing, keep exploring, and you'll become pros at this in no time. Thanks for joining me on this physics adventure!