Displacement Calculation: Constant Acceleration On The Ox Axis

Hey guys! Let's dive into a physics problem! We're talking about a body moving along the Ox axis with constant acceleration. We've got some initial conditions and we need to figure out the displacement. Sounds fun, right?

Understanding the Problem: Constant Acceleration Motion

So, the core of this problem is understanding motion with constant acceleration. What does that mean? Basically, the object's velocity is changing at a steady rate. This is super common in the real world – think about a car speeding up, or a ball rolling down a hill (ignoring air resistance, of course!). We're given the initial velocity ($U_{0x}$), the final velocity ($U_x$), and the acceleration ($a_x$). Our mission? To find the displacement, which is how far the object moved along the Ox axis.

Now, before we get all equations-y, let's make sure we've got the basics down. Acceleration is the rate of change of velocity. If an object is speeding up, it has positive acceleration. If it's slowing down, it has negative acceleration (also called deceleration). The displacement is a vector quantity, meaning it has both magnitude (how much) and direction (which way). In this case, since we're only dealing with motion along the Ox axis, the direction is straightforward: along the axis.

Let's break down the information we have, shall we? The initial velocity ($U_{0x}$) is -3.5 m/s. The minus sign tells us the object is initially moving in the negative direction of the Ox axis. The final velocity ($U_x$) is 1.5 m/s. This means the object has slowed down, stopped, and then started moving in the positive direction of the Ox axis. Finally, the acceleration ($a_x$) is 2.5 m/s². The positive value means the object is accelerating in the positive direction. Got it?

To make sure we're all on the same page, let's talk about the key concepts involved. Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. We're using kinematic equations here. These equations relate displacement, initial velocity, final velocity, acceleration, and time. And the cool thing is, we've got the equations to solve this problem! It's like having a secret code to unlock the answer!

This kind of problem is super fundamental in physics, forming the basis for more complex concepts like projectile motion and dynamics. This is why it's so important to have a solid understanding of how to work through these types of questions. We'll use this knowledge to solve problems, understand the physical world around us, and maybe even impress our friends with our physics prowess. Let's get started. We've got this!

Applying the Kinematic Equation for Displacement

Alright, time to get to the juicy part – solving the problem! We need to find the displacement, often denoted as $S_x$, given $U_{0x}$, $U_x$, and $a_x$. We're lucky enough to have a kinematic equation that directly relates these quantities. The equation we'll use is: $U_x^2 = U_{0x}^2 + 2 * a_x * S_x$.

Let's unpack this equation. $U_x$ is the final velocity, $U_{0x}$ is the initial velocity, $a_x$ is the acceleration, and $S_x$ is the displacement we're trying to find. See how it all fits together? It's like a puzzle, and we have all the pieces except for one. Now we can rearrange the equation to solve for $S_x$. First, subtract $U_{0x}^2$ from both sides: $U_x^2 - U_{0x}^2 = 2 * a_x * S_x$. Then, divide both sides by $2 * a_x$: $(U_x^2 - U_{0x}^2) / (2 * a_x) = S_x$. This is our formula for the displacement.

Now, it's plugging in the numbers! We know:

• $U_{0x} = -3.5 ext{ m/s}$
• $U_x = 1.5 ext{ m/s}$
• $a_x = 2.5 ext{ m/s}^2$

Substituting these values into the equation, we get:

$S_x = (1.5^2 - (-3.5)^2) / (2 * 2.5)$

Let's do the math carefully. First, square the velocities: $1.5^2 = 2.25$ and $(-3.5)^2 = 12.25$. Then, subtract: $2.25 - 12.25 = -10$. Finally, divide by $2 * 2.5 = 5$: $-10 / 5 = -2$. So, the displacement, $S_x$, is -2 meters.

This negative sign is important! It tells us that the displacement is in the negative direction of the Ox axis. It means the object moved 2 meters in the negative direction while its velocity changed from -3.5 m/s to 1.5 m/s under the constant acceleration of 2.5 m/s². The negative sign just tells us the direction of the movement relative to our chosen coordinate system.

Remember, in physics, it's super important to keep track of the units. In this case, our displacement is in meters (m), which is the standard unit of length in the International System of Units (SI). Always check that your units are consistent throughout the problem and that your final answer has the correct units.

Interpreting the Results and Further Analysis

Okay, so we've found that the displacement of the body is -2 meters. What does this really mean in terms of the body's motion? Well, it tells us that, overall, the body moved 2 meters in the negative direction along the Ox axis. Even though the body initially had a negative velocity (moving in the negative direction), and then the acceleration changed the body's direction, the net displacement was still in the negative direction.

Think about it this way: the body started moving left (negative direction), then slowed down, stopped, and then started moving right (positive direction). However, it didn't travel far enough to