Particle Motion: Time To Rest With Variable Acceleration

Oct 28, 2025 by Admin 57 views

Hey guys! Let's dive into a classic physics problem involving particle motion and variable acceleration. This is the kind of question that might seem tricky at first, but once you break it down, it's totally manageable. We're going to tackle a scenario where a particle starts from rest and has an acceleration that changes over time. Our mission? To figure out when this particle will momentarily come to rest again. So, buckle up, and let's get started!

Understanding the Problem

Okay, so here's the deal: we've got a particle chilling at rest, and then, bam! It starts accelerating. But this isn't your regular constant acceleration; it's a time-dependent acceleration, given by the equation a(t) = 6 - 3t (in m/s²). This means the acceleration changes as time (t) goes on. The key question we need to answer is: At what later time does this particle momentarily come to rest again?

To really nail this, we need to understand what's happening physically. The particle starts at rest, accelerates, gains speed, and then, at some point, it starts slowing down until it momentarily stops before potentially changing direction. That "momentarily" part is crucial because it tells us we're looking for a specific instant in time when the particle's velocity is zero. This is a common concept in physics, and mastering it is super important for solving more complex problems later on. Think of it like throwing a ball straight up in the air. It slows down, stops for a split second at its highest point, and then starts falling back down. We're looking for that split second of zero velocity in our particle's journey.

Now, let's break down the information we have. We know the acceleration as a function of time: a(t) = 6 - 3t. We also know the initial condition: the particle starts from rest. This means its initial velocity, v(0), is 0 m/s. What we're trying to find is the time t when the velocity, v(t), is also 0 m/s again. This is where our physics toolbox comes in handy. We need to relate acceleration, velocity, and time, and that's where the concepts of integration and kinematics come into play.

Connecting Acceleration, Velocity, and Time

Alright, so how do we connect acceleration, velocity, and time? This is where the fundamental relationship between them comes into play. Remember, acceleration is the rate of change of velocity with respect to time. Mathematically, this is expressed as:

a(t) = dv(t)/dt

Where:

a(t) is the acceleration at time t.
dv(t) is the change in velocity.
dt is the change in time.

This equation is super important because it tells us that if we know the acceleration as a function of time, we can find the velocity by integrating the acceleration with respect to time. Think of it like this: if acceleration is the speeding up or slowing down rate, then velocity is the cumulative effect of that speeding up or slowing down over time. To get the cumulative effect, we need to integrate. So, we're going to integrate both sides of the equation:

∫dv(t) = ∫a(t) dt

This gives us:

v(t) = ∫a(t) dt

Now, let's plug in our acceleration function, a(t) = 6 - 3t:

v(t) = ∫(6 - 3t) dt

Time to put on our calculus hats! Integrating this expression is pretty straightforward. Remember the power rule for integration: ∫xⁿ dx = (xⁿ⁺¹)/(n+1) + C, where C is the constant of integration. Applying this rule, we get:

v(t) = 6t - (3/2)t² + C

Don't forget that constant of integration, C! It's crucial because it represents the initial velocity. Remember, we were told that the particle starts from rest, meaning v(0) = 0. We can use this information to find the value of C. Let's plug in t = 0 and v(t) = 0:

0 = 6(0) - (3/2)(0)² + C

This simplifies to:

C = 0

So, our velocity function becomes:

v(t) = 6t - (3/2)t²

We're getting closer! We now have an equation that tells us the velocity of the particle at any time t. Our original question was to find the time when the particle momentarily comes to rest again, meaning when v(t) = 0. Let's use this condition to solve for t.

Finding the Time When the Particle Comes to Rest

Okay, we've got our velocity function: v(t) = 6t - (3/2)t². Now, we want to find the time t when the particle momentarily comes to rest, which means v(t) = 0. So, let's set our equation equal to zero:

0 = 6t - (3/2)t²

This is a quadratic equation, and we can solve it for t. First, let's make things a bit easier by factoring out a common factor. We can factor out a t from both terms:

0 = t(6 - (3/2)t)

Now we have a product of two factors that equals zero. This means that either the first factor, t, is zero, or the second factor, (6 - (3/2)t), is zero. Let's consider each case:

Case 1: t = 0

This solution corresponds to the initial condition, which we already knew. At t = 0, the particle is at rest. But we're interested in the later time when it comes to rest again, so this solution isn't what we're looking for.

Case 2: 6 - (3/2)t = 0

This is the interesting case! Let's solve this equation for t. First, add (3/2)t to both sides:

6 = (3/2)t

Now, multiply both sides by (2/3) to isolate t:

t = 6 * (2/3)

t = 4

So, we've found our answer! The particle momentarily comes to rest again at t = 4 seconds.

Conclusion

Woohoo! We did it! By understanding the relationship between acceleration, velocity, and time, and using a little calculus magic, we figured out that the particle will momentarily come to rest again at t = 4 seconds. This problem highlights the power of calculus in solving physics problems, especially those involving variable acceleration. Remember, breaking down the problem into smaller steps, understanding the concepts, and applying the right tools are key to success. Keep practicing, and you'll be a physics whiz in no time! Now go out there and conquer some more physics challenges, guys! You've got this! 🚀✨