Solving Motion Equations Trajectory, Velocity, And Acceleration

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Hey guys! Let's dive into the fascinating world of motion equations and how to solve them. In this article, we're going to tackle a classic problem in physics: finding the trajectory, velocity, and acceleration of a material point given its equation of motion. It might sound intimidating, but trust me, we'll break it down step by step so it's super easy to understand. So, buckle up and let's get started!

Understanding the Problem

Before we jump into the calculations, let's make sure we understand the problem statement. We're given the equation of motion of a material point as r(t) = (t^2 / 2) i + t j. This equation tells us the position of the point in space at any given time t. Here, i and j are unit vectors in the x and y directions, respectively. Essentially, this equation describes a curve in a two-dimensional plane.

Our mission, should we choose to accept it, is threefold:

  • a. Find the equation of the trajectory: This means we need to find a relationship between the x and y coordinates of the point that doesn't involve time t. Think of it as tracing the path the point takes as it moves.
  • b. Find the magnitude of the velocity at t = 2 s: Velocity is the rate of change of position with respect to time. So, we'll need to find the velocity vector first and then calculate its magnitude (speed) at the specific time t = 2 s.
  • c. Find the instantaneous acceleration in the time interval t ∈ (0, 4) s: Acceleration is the rate of change of velocity with respect to time. We'll need to find the acceleration vector as a function of time within the given interval.

Key Concepts to Remember

To solve this problem effectively, it's crucial to have a solid grasp of a few key concepts from physics and calculus. Don't worry, we'll review them briefly:

  • Position Vector (r(t)): As we've seen, this vector describes the location of the particle at a given time t. It's often expressed in terms of its components along the coordinate axes (e.g., x(t)i + y(t)j).
  • Velocity Vector (v(t)): This is the time derivative of the position vector, v(t) = dr(t)/dt. It tells us how fast the particle is moving and in what direction.
  • Acceleration Vector (a(t)): This is the time derivative of the velocity vector, a(t) = dv(t)/dt. It tells us how the velocity is changing with time.
  • Trajectory: This is the path that the particle follows in space. Mathematically, it's the relationship between the coordinates (x, y, z, etc.) without involving time t.
  • Magnitude of a Vector: For a vector A = Axi + Ayj, the magnitude is given by |A| = √(Ax^2 + Ay^2).

With these concepts in mind, we're well-equipped to tackle the problem. Let's dive into the solutions!

Solving for the Trajectory (Part a)

First, let's find the equation of the trajectory. Remember, the trajectory is the path the particle takes, and we want to express this path as a relationship between the x and y coordinates without involving time t. The given equation of motion is:

r(t) = (t^2 / 2) i + t j

This equation gives us the x and y components of the position as functions of time:

  • x(t) = t^2 / 2
  • y(t) = t

Our goal is to eliminate t from these equations. We can easily do this by solving the second equation for t:

t = y

Now, substitute this expression for t into the first equation:

x = (y^2) / 2

Multiplying both sides by 2, we get:

2x = y^2

This is the equation of the trajectory. It represents a parabola opening along the positive x-axis. So, the path the material point follows is a parabolic curve. Great job! We've solved the first part of the problem.

Finding the Velocity at t = 2 s (Part b)

Next up, we need to find the magnitude of the velocity of the material point at time t = 2 s. Remember that velocity is the rate of change of position with respect to time. Mathematically, this means we need to find the derivative of the position vector r(t) with respect to t.

So, let's differentiate r(t):

r(t) = (t^2 / 2) i + t j

v(t) = dr(t)/dt = d/dt [(t^2 / 2) i + t j]

Differentiating component-wise, we get:

v(t) = (t) i + (1) j

This is the velocity vector as a function of time. Now, we need to find the velocity at t = 2 s. Plug in t = 2 into the equation:

v(2) = (2) i + j

So, the velocity vector at t = 2 s is 2i + j. But we're not done yet! The question asks for the magnitude of the velocity. To find the magnitude, we use the formula:

|v| = √(vx^2 + vy^2)

In our case, vx = 2 and vy = 1. So,

|v(2)| = √(2^2 + 1^2) = √(4 + 1) = √5

Therefore, the magnitude of the velocity of the material point at t = 2 s is √5 units (assuming we're working in standard units like meters and seconds). Awesome! We've conquered the second part of the problem.

Determining the Instantaneous Acceleration (Part c)

Finally, let's tackle the third part of the problem: finding the instantaneous acceleration in the time interval t ∈ (0, 4) s. Remember that acceleration is the rate of change of velocity with respect to time. So, we need to find the derivative of the velocity vector v(t) with respect to t.

We already found the velocity vector in the previous part:

v(t) = t i + j

Now, let's differentiate v(t):

a(t) = dv(t)/dt = d/dt [t i + j]

Differentiating component-wise, we get:

a(t) = (1) i + (0) j

a(t) = i

This is the acceleration vector. Notice that it's constant and doesn't depend on time t. The acceleration is simply i, which means it's a constant acceleration of 1 unit in the positive x-direction. This holds true for the entire time interval t ∈ (0, 4) s. Fantastic! We've solved the final part of the problem.

Conclusion

Alright, guys, we did it! We successfully found the equation of the trajectory, the magnitude of the velocity at a specific time, and the instantaneous acceleration for the given equation of motion. Let's recap our findings:

  • a. Equation of the trajectory: 2x = y^2 (a parabola)
  • b. Magnitude of the velocity at t = 2 s: √5 units
  • c. Instantaneous acceleration in t ∈ (0, 4) s: a(t) = i (constant acceleration in the x-direction)

By breaking down the problem into smaller, manageable steps and applying the fundamental concepts of kinematics and calculus, we were able to solve it with confidence. Remember, the key to mastering physics is to understand the underlying principles and practice applying them to different scenarios. So, keep exploring, keep learning, and keep those equations of motion rolling!

I hope this article helped you understand how to solve motion equations. If you have any questions or want to explore more physics problems, feel free to ask. Keep up the great work, and I'll catch you in the next one!