Calculating Initial Velocity In Accelerated Linear Motion

Hey guys! Let's dive into a classic physics problem. We're gonna figure out how to calculate the initial velocity of an object undergoing uniformly accelerated linear motion. The problem gives us some key info: the distance traveled, the change in velocity, and the acceleration. Sounds fun, right? Don't worry, it's not as scary as it seems! We will break it down step-by-step. Get ready to flex those physics muscles! Understanding this is crucial for anyone studying physics because it forms the bedrock for understanding more complex motion scenarios. This concept appears regularly in various examinations, making a solid comprehension very beneficial for your academic performance. In this article, we'll not only solve the problem, but also explore the underlying concepts and formulas, ensuring you have a firm grasp of the topic. So, let’s get started and make physics a bit more approachable and enjoyable! Let’s get to work!

Understanding the Problem and Given Information

Alright, let's unpack this problem together. We're dealing with uniformly accelerated linear motion. This means the object is moving in a straight line, and its velocity is changing at a constant rate. That constant rate is what we call acceleration. Now, let’s list what we know. First, the object covers a distance, $S$, of 0.03 km. But, we need to convert that into meters (m) because we're working with acceleration in m/s². So, $S = 0.03 ext{ km} = 30 ext{ m}$. Cool? Second, the problem states that the magnitude (or, the size) of the velocity increases fourfold. Let's denote the initial velocity as $v_0$ and the final velocity as $v$. This means that $v = 4v_0$. The third piece of info is the acceleration, $a = 1 ext{ m/s}^2$. The question we need to answer is: What is the initial velocity, $v_0$? The key is choosing the right kinematic equation. Remember, kinematic equations are your best friends in these types of problems. They give you the relationship between displacement, velocity, acceleration, and time. With these, you can solve many motion problems.

The Kinematic Equations: Your Physics Toolkit

There are several kinematic equations that we can use, but the one that fits our needs perfectly here doesn't involve time. Time is something that we don't know directly from the problem. So, here's the equation we'll use:

$v^2 = v_0^2 + 2aS$

Where:

• $v$ is the final velocity
• $v_0$ is the initial velocity
• $a$ is the acceleration
• $S$ is the displacement.

This equation is super useful because it directly relates all the quantities we know ($S$ and $a$) and the two velocities ($v$ and $v_0$). Now, we can substitute the known values into the equation and solve for the initial velocity, $v_0$. We also know that $v = 4v_0$, which helps simplify things. This is the beauty of choosing the right formula; it simplifies calculations and helps us reach the correct answer faster and with more clarity. Remember, these equations are powerful tools; master them, and you’ll ace your physics problems! The key to success is practice, practice, and more practice! Let's get to work!

Solving for the Initial Velocity

Okay, let's put our knowledge into practice and calculate that initial velocity! Remember, we have the equation: $v^2 = v_0^2 + 2aS$. And we also know that $v = 4v_0$. Now, let’s substitute $v$ with $4v_0$ into our main equation. This gives us:

$(4v_0)^2 = v_0^2 + 2aS$

Simplifying further, we get:

$16v_0^2 = v_0^2 + 2aS$

Now, let's plug in the values for $a$ and $S$: $a = 1 ext{ m/s}^2$ and $S = 30 ext{ m}$.

$16v_0^2 = v_0^2 + 2 * 1 * 30$

Which simplifies to:

$16v_0^2 = v_0^2 + 60$

Now, we need to isolate $v_0$. Subtract $v_0^2$ from both sides of the equation:

$15v_0^2 = 60$

Divide both sides by 15:

$v_0^2 = 4$

Finally, take the square root of both sides to solve for $v_0$: $v_0 = ext{±} 2 ext{ m/s}$. Since we are looking for the magnitude of the initial velocity, we can disregard the negative sign. So, the initial velocity, $v_0$, is 2 m/s. This is an exciting step because now we are near the solution. By following each step, we gradually arrive at the solution. Physics problems are really like solving puzzles, aren’t they? This approach not only provides the answer but also enhances your understanding of how to apply kinematic equations in similar situations.

Choosing the Correct Answer from the Options

Great job, we’ve found the answer! Looking back at the options provided in the problem, we see that 2 m/s is indeed one of the choices. Thus, the correct answer is 1) 2 m/s. Easy peasy, right? The calculations were straightforward because we understood the fundamentals. So, in these types of physics problems, the key is to choose the right equation. You have to understand what you know and what you're trying to find. This logical approach will allow you to solve almost any problem. The more you practice, the easier it becomes. Now, let’s recap what we've learned and wrap this up!

Conclusion: Mastering Uniformly Accelerated Motion

Awesome work, everyone! We've successfully navigated a physics problem involving uniformly accelerated linear motion. We started with the basics, reviewed the key concepts, selected the appropriate kinematic equation, and meticulously solved for the initial velocity. Remember that the process involves understanding the problem, identifying the given parameters, and correctly applying the appropriate formula. This approach is not only applicable to this specific problem but also to a wide range of physics scenarios. Keep in mind: practice is key! By working through similar problems, you'll gain confidence and sharpen your problem-solving skills. Physics can be a lot of fun, and it can also teach you about the world around you. This entire process builds your confidence and improves your problem-solving skills, and that is what we want! Every time you solve a physics problem, you get a little smarter. So, keep it up, keep learning, and keep asking questions. If you are struggling with a similar problem, go back through the steps; don’t give up. The most important thing is to understand the concepts. Now go out there and conquer more physics problems!