Particle Motion: Time, Position, Acceleration At V=0

Hey guys! Ever found yourself scratching your head over particle motion problems in physics? You're not alone! These problems can seem tricky, but with the right approach, they become super manageable. Today, we're diving deep into a classic particle motion problem, breaking it down step-by-step so you can conquer similar challenges with confidence. We'll be tackling a scenario where we need to determine the time, position, and acceleration of a particle when its velocity hits zero. Buckle up, let's get started!

The Problem: A Particle's Journey

Let's jump straight into the problem. We're given the relationship that defines the motion of a particle as 2 = 33 - 512 - 30t + 8x, where 'x' represents the position in feet and 't' represents the time in seconds. Our mission, should we choose to accept it (and we do!), is to find:

• The time (t) when the velocity (V) is zero.
• The position (x) at that moment.
• The acceleration (a) at that very instant.

This might seem like a lot, but don't worry! We'll break it down into bite-sized pieces and solve it together. Think of it like a puzzle – each piece of information we find helps us fit the bigger picture.

Step 1: Unveiling the Velocity

Our first order of business is to find the velocity of the particle. Remember, velocity is the rate of change of position with respect to time. In calculus terms, this means we need to differentiate the position function with respect to time. But hold on! Our equation isn't directly in the form of x(t). It's a bit… disguised. We need to do a little algebraic maneuvering first to isolate 'x'.

Let's rewrite the equation to solve for x:

8x = 512 + 30t - 33

8x = 479 + 30t

x = (479/8) + (30/8)t

Now we have x as a function of t, which is exactly what we need! We can simplify it further:

x(t) = 59.875 + 3.75t

Now, the fun part begins! We'll differentiate x(t) with respect to t to find the velocity function, v(t). Remember, the derivative of a constant is zero, and the derivative of 'at' (where 'a' is a constant) is simply 'a'. So:

v(t) = dx/dt = 3.75 ft/s

Wait a minute… This is interesting! The velocity turned out to be a constant value. It doesn't depend on time. This means the particle is moving at a constant velocity of 3.75 ft/s. So, the question of finding the time when V=0 seems a bit odd in this context. However, let's revisit the original equation to ensure we haven't missed anything crucial in our interpretation or calculation.

A Closer Look at the Original Equation

The initial equation given was 2 = 33 - 512 - 30t + 8x. There seems to be a typo or an incomplete equation. It's likely that the "2 =" part was intended to be part of a larger term, perhaps involving 'x' or 't'. Without the correct equation, proceeding to find when V=0 is impossible, because the derivative was calculated from an incorrect premise. Let’s assume that the equation was intended to be a quadratic one, which is more typical for these types of problems. A more likely form of the equation could be x² = 33 - 5t² - 30t + 8x. If we proceeded with this assumption, we’d have a much more complex (and typical) motion problem.

Assuming a Quadratic Equation: x² = 33 - 5t² - 30t + 8x

Let’s work through the problem with this new assumption to see how it changes the solution. First, we need to rearrange the equation:

x² - 8x = 33 - 5t² - 30t

To make it easier to differentiate implicitly, we can differentiate both sides with respect to time (t):

d/dt (x² - 8x) = d/dt (33 - 5t² - 30t)

Using the chain rule, we get:

(2x * dx/dt) - (8 * dx/dt) = -10t - 30

Now, remember that dx/dt is the velocity, v. So, we can rewrite the equation as:

(2x - 8)v = -10t - 30

Now we have a relationship between x, v, and t. To find the velocity, we need to differentiate the initial equation implicitly with respect to t.

Differentiating Again for Acceleration

To find the acceleration, we’ll need to differentiate our velocity equation with respect to time. This will get a bit more complex, but let’s take it step by step. First, let's express the velocity explicitly. However, to get v as a direct function of t, it's evident we will need the original x in terms of t, so let's go back to the adjusted position equation:

x² - 8x + 5t² + 30t - 33 = 0

This is a quadratic equation in terms of x. We can solve for x using the quadratic formula. Once we have x(t), we can find v(t) by differentiating, and then a(t) by differentiating v(t).

Solving the Quadratic for x(t)

The quadratic formula is: x = [-b ± sqrt(b² - 4ac)] / (2a). In our case, a = 1, b = -8, and c = 5t² + 30t - 33. Plugging these in:

x(t) = [8 ± sqrt((-8)² - 4 * 1 * (5t² + 30t - 33))] / (2 * 1)

x(t) = [8 ± sqrt(64 - 20t² - 120t + 132)] / 2

x(t) = [8 ± sqrt(-20t² - 120t + 196)] / 2

x(t) = 4 ± sqrt(-5t² - 30t + 49)

Now we have x(t), but it’s quite complex! This will lead to a complicated expression for velocity and acceleration. However, this more complex form is typical for physics problems involving variable acceleration.

Step 2: Finding the Time When Velocity is Zero (V=0)

Now that we (attempted to) find the velocity function, we need to determine the time(s) when v(t) = 0. If we stick to our original (likely incorrect) interpretation, the velocity was constant (3.75 ft/s), so it never equals zero. This suggests that the original equation might indeed be flawed or incomplete.

However, using our adjusted quadratic interpretation, finding when v(t) = 0 would involve differentiating the complicated x(t) we derived, setting it to zero, and solving for t. This is a significant algebraic challenge! Typically, for exam-like problems, these equations simplify nicely, so the complexity here reinforces the possibility of a different intended equation or setup.

Step 3: Determining Position and Acceleration

Once we find the time(s) when the velocity is zero, we can plug those time values back into the position function x(t) to find the position of the particle at those moments. Similarly, we'd plug the time values into the acceleration function a(t) to find the acceleration.

However, given the ambiguity in the original equation and the complexity arising from our quadratic assumption, finding these values precisely is not feasible without further clarification of the initial problem statement.

Key Takeaways and Lessons Learned

This problem, while seemingly straightforward at first glance, highlights the critical importance of:

• Accurate Problem Statements: A slight ambiguity or typo in the given information can drastically change the nature and complexity of the solution.
• Understanding the Physics: Recognizing that velocity is the derivative of position and acceleration is the derivative of velocity is fundamental.
• Algebraic Dexterity: Manipulating equations, solving for variables, and using the chain rule are essential skills.
• Interpreting Results: Sometimes, the mathematical results can indicate an issue with the problem setup, requiring a reevaluation of assumptions.

In conclusion, while we couldn't arrive at a complete numerical solution due to the likely typo in the initial equation, we've thoroughly explored the process of solving particle motion problems. Remember, guys, practice makes perfect! The more you work through these types of problems, the more comfortable you'll become with the concepts and techniques involved.

If you encounter a similar problem, double-check the given information, break the problem down into smaller steps, and don't be afraid to revisit your assumptions if the results don't seem to make sense. Keep learning, keep practicing, and you'll master particle motion in no time!