Particle Position: Solving For Displacement At T=4 Seconds

Hey guys! Let's dive into a classic physics problem involving particle motion. We've got a particle zooming along a straight line, and its velocity is described by the equation v = (4t - 3t^2). The goal? To figure out the particle's position at t = 4 seconds, knowing that it started at s = 0 when t = 0. Sounds like a fun challenge, right? So, buckle up, and let's get started!

Understanding the Problem: Velocity, Position, and Time

Before we jump into the math, let's make sure we're all on the same page with the concepts. In this problem, the key idea is the relationship between velocity, position, and time. Velocity tells us how fast the particle is moving and in what direction. Position, on the other hand, tells us where the particle is located at a specific time. And, of course, time is the independent variable that governs the motion.

In calculus terms, velocity is the derivative of position with respect to time. This means that if we know the position function, we can find the velocity function by differentiating it. Conversely, if we know the velocity function, we can find the position function by integrating it. This is exactly what we'll be doing in this problem!

Why is this so important? Well, in many real-world scenarios, we often have information about an object's velocity or acceleration, but we want to know its position. Think about tracking a car's movement using its speedometer readings or predicting the trajectory of a rocket based on its engine thrust. Understanding the relationship between velocity, position, and time is crucial for solving these kinds of problems.

Moreover, the problem gives us an initial condition: s = 0 when t = 0. This is vital because when we integrate the velocity function to find the position function, we'll get a constant of integration. The initial condition allows us to determine the value of this constant and find the unique position function that describes the particle's motion. So, keep this in mind as we move forward!

Setting up the Integral: From Velocity to Position

The first key step in solving this problem is to recognize that we need to find the position function, s(t), given the velocity function, v(t) = 4t - 3t^2. As we discussed earlier, position is the integral of velocity with respect to time. So, we can write:

s(t) = ∫ v(t) dt = ∫ (4t - 3t^2) dt

This integral represents the area under the velocity curve. Remember, integration is like the reverse of differentiation. We're essentially asking: what function, when differentiated, gives us 4t - 3t^2? To solve this, we'll use the power rule for integration, which states that:

∫ t^n dt = (t^(n+1)) / (n+1) + C

where C is the constant of integration. This constant is super important, guys, because it represents the initial position of the particle. Without it, we'd have a whole family of possible position functions, and we wouldn't know which one is the actual position function for our particle. This is where the initial condition we talked about earlier comes into play!

Now, let's apply the power rule to our integral. We'll integrate each term in the velocity function separately. First, we integrate 4t: ∫ 4t dt = 4 ∫ t^1 dt = 4 * (t^2 / 2) = 2t^2. Next, we integrate -3t^2: ∫ -3t^2 dt = -3 ∫ t^2 dt = -3 * (t^3 / 3) = -t^3. Don't forget to add the constant of integration, C!

So, after integrating, we have:

s(t) = 2t^2 - t^3 + C

This is our general position function. It tells us the particle's position at any time t, but we still need to find the value of C. That's where the next step comes in!

Applying the Initial Condition: Finding the Constant of Integration

Alright, we've got the general form of the position function: s(t) = 2t^2 - t^3 + C. Now, let's use the initial condition that was provided in the problem to pinpoint the exact value of C. Remember, we were told that s = 0 when t = 0. This is our anchor point, the one piece of information that ties down our position function.

What we're going to do is substitute these values into our general position function. We'll plug in t = 0 and s(0) = 0 into the equation and solve for C. It's like fitting a puzzle piece into place! So, let's do it:

s(0) = 2(0)^2 - (0)^3 + C = 0

Simplifying this, we get:

0 = 0 - 0 + C

And just like that, we find that C = 0! This means that our initial position is indeed at the origin, which makes sense given the problem statement. Now, we have the complete position function:

s(t) = 2t^2 - t^3

This function tells us exactly where the particle is at any time t. It's like having a GPS for our particle! But we're not quite done yet. The problem asks us to find the position at a specific time, t = 4 seconds. So, let's move on to the final step and plug in that value!

Calculating the Position at t=4 Seconds: The Final Answer

We've made it to the final stretch, guys! We've got the position function, s(t) = 2t^2 - t^3, and we know we need to find the position at t = 4 seconds. This is the home stretch, where we get to plug in the numbers and get our final answer. It's like the satisfying click of the last puzzle piece falling into place!

To find the position at t = 4 seconds, we simply substitute t = 4 into our position function:

s(4) = 2(4)^2 - (4)^3

Now, let's do the math. First, we calculate 4^2, which is 16. Then, we multiply that by 2, giving us 32. Next, we calculate 4^3, which is 64. So, we have:

s(4) = 32 - 64

Finally, we subtract 64 from 32, which gives us -32. Therefore, the position of the particle at t = 4 seconds is:

s(4) = -32

But wait! We need to include the units in our answer. The problem states that time is in seconds, but what about position? Since the velocity is given without explicit units, we can assume the position will be in the same units as the displacement implied by the velocity. If the velocity was in meters per second (m/s), then the position would be in meters (m). So, our final answer is:

s(4) = -32 meters

And there we have it! The particle is located at -32 meters at t = 4 seconds. The negative sign indicates that the particle is 32 meters to the left of its starting position (assuming the positive direction is to the right). We solved it! Give yourselves a pat on the back.

Conclusion: Mastering Motion Problems

Awesome job, guys! We've successfully tackled this particle motion problem. We started with the velocity function, used integration to find the position function, applied the initial condition to determine the constant of integration, and finally, calculated the position at t = 4 seconds. That's a lot of physics in one problem!

The key takeaways from this exercise are the fundamental relationships between velocity, position, and time, and how calculus, specifically integration, helps us move between these concepts. Remember, velocity is the derivative of position, and position is the integral of velocity. These are powerful tools in physics, and mastering them opens the door to understanding more complex motion problems.

Don't forget the importance of initial conditions. They are crucial for finding the specific solution to a problem, especially when dealing with integration. The constant of integration is like a missing piece of the puzzle, and initial conditions help us find that piece.

Keep practicing these types of problems, and you'll become a pro at solving motion problems. Physics is all about understanding the world around us, and these concepts are fundamental to that understanding. So, keep exploring, keep questioning, and keep learning!

If you have any questions or want to dive deeper into this topic, feel free to ask! Until next time, keep moving forward!