Motion Problem: Position And Time Calculation

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Motion Problem: Position and Time Calculation

Hey guys! Let's dive into a classic motion problem where we'll figure out the position of a moving object at a specific time and also determine when it reaches a certain position. This is a super common type of problem in physics, and once you get the hang of it, you'll be solving these like a pro. We're going to break down each step so it's crystal clear.

Understanding the Problem

Okay, so we have a mobile object moving along a straight line. The key here is the time function given: s = 6 - 5t + t². This equation tells us the position (s) of the object at any given time (t). The units are in SI, which means s is in meters and t is in seconds. We have two main questions to answer:

  1. What is the object's position at t = 5 seconds?
  2. At what time does the object pass the position s = -50 meters?

Part A: Finding the Position at t = 5s

This part is pretty straightforward. We're given the time (t = 5s) and the equation for position (s). All we need to do is plug in the value of t into the equation and solve for s. This is where our algebra skills come into play!

So, let's take the equation: s = 6 - 5t + t²

Now, substitute t with 5:

s = 6 - 5(5) + (5)²

Let's simplify this step-by-step:

s = 6 - 25 + 25

Notice that -25 and +25 cancel each other out:

s = 6

So, at t = 5 seconds, the position of the object is 6 meters. Easy peasy, right? We've found our first answer. Now, let's move on to the second part, which is a bit more interesting.

Part C: Finding the Time When s = -50m

This part requires a bit more algebraic manipulation. We're given the position (s = -50m) and we need to find the time (t) when the object is at that position. Again, we'll use the same equation: s = 6 - 5t + t².

This time, we'll substitute s with -50:

-50 = 6 - 5t + t²

Now, we need to rearrange this equation into a standard quadratic form, which is at² + bt + c = 0. To do this, let's add 50 to both sides of the equation:

0 = t² - 5t + 6 + 50

Simplify it:

0 = t² - 5t + 56

Now we have a quadratic equation. There are a few ways to solve this, but the most common method is using the quadratic formula. Remember that? It's a lifesaver for these types of problems!

Solving the Quadratic Equation

The quadratic formula is: t = [-b ± √(b² - 4ac)] / 2a

Where a, b, and c are the coefficients from our quadratic equation at² + bt + c = 0. In our case:

  • a = 1 (the coefficient of t²)
  • b = -5 (the coefficient of t)
  • c = 56 (the constant term)

Let's plug these values into the quadratic formula:

t = [ -(-5) ± √((-5)² - 4 * 1 * 56) ] / (2 * 1)

Simplify it:

t = [ 5 ± √(25 - 224) ] / 2

t = [ 5 ± √(-199) ] / 2

Uh oh! We've run into a problem. We have a negative number inside the square root. This means that the solutions for t will be imaginary numbers. In the context of this problem, this implies that the object never actually reaches the position s = -50 meters. It's important to recognize these situations when dealing with physics problems.

Interpreting the Result

The fact that we got imaginary solutions for t tells us something important about the motion of the object. It means that based on the given equation of motion, the object never reaches the position -50 meters. It might slow down, change direction, or simply not travel far enough to reach that point. This is a crucial part of problem-solving – not just finding the answer, but also understanding what the answer (or lack thereof) means in the real world.

Key Takeaways

Let's recap what we've learned from this problem:

  • Understanding the equation of motion: The equation s = 6 - 5t + t² is the foundation. It tells us how the object's position changes with time.
  • Substituting values: We can find the position at a specific time by simply substituting the time value into the equation.
  • Quadratic equations: Finding the time at a specific position might involve solving a quadratic equation. Remember the quadratic formula!
  • Interpreting results: It's crucial to understand what the solutions mean in the context of the problem. Imaginary solutions can indicate that the condition is never met.

Practice Makes Perfect

The best way to get comfortable with these types of problems is to practice! Try working through similar examples, changing the values, and seeing how the results change. You can also try graphing the equation of motion to visualize the object's movement. This can give you a better intuitive understanding of what's going on.

Conclusion

So, there you have it! We've solved a motion problem by finding the position at a given time and attempting to find the time at a given position. Remember to break down the problem into steps, use the appropriate formulas, and most importantly, interpret your results. Keep practicing, and you'll become a motion problem-solving master in no time! If you guys have any questions, feel free to ask. Let's keep learning together!

Now that we've tackled this problem, let's delve deeper into some related concepts and explore variations of this type of question. Understanding the underlying principles will make you even more confident in solving motion problems.

Diving Deeper into Kinematics

This problem is rooted in the field of kinematics, which is a branch of physics that describes the motion of objects without considering the forces that cause the motion. In other words, we're focusing on how things move, not why they move. Key concepts in kinematics include:

  • Displacement: The change in position of an object.
  • Velocity: The rate of change of displacement with respect to time.
  • Acceleration: The rate of change of velocity with respect to time.

Our equation s = 6 - 5t + t² describes the displacement of the object as a function of time. To find the velocity, we would need to take the derivative of this equation with respect to time. Similarly, to find the acceleration, we would take the derivative of the velocity equation.

Velocity and Acceleration

Let's find the velocity and acceleration for our object. If s = 6 - 5t + t², then the velocity (v) is the derivative of s with respect to t:

v = ds/dt = -5 + 2t

This equation tells us how the velocity of the object changes over time. Now, let's find the acceleration (a), which is the derivative of v with respect to t:

a = dv/dt = 2

Notice that the acceleration is constant and equal to 2 m/s². This means the object is undergoing uniform acceleration. Understanding the acceleration can give us valuable insights into the object's motion.

Graphical Representation

Another way to visualize the motion is by graphing the position, velocity, and acceleration as functions of time. Here's a brief overview of what these graphs would look like:

  • Position vs. Time (s vs. t): This would be a parabola, since our position equation is a quadratic function. The vertex of the parabola would represent the point where the object changes direction.
  • Velocity vs. Time (v vs. t): This would be a straight line, since our velocity equation is a linear function. The slope of the line would represent the acceleration.
  • Acceleration vs. Time (a vs. t): This would be a horizontal line, since the acceleration is constant. The y-value of the line would represent the magnitude of the acceleration.

Visualizing these graphs can be a powerful tool for understanding the motion described by the equations.

Variations of the Problem

Now, let's think about some variations of this problem that you might encounter:

  • Initial conditions: The problem might give you initial conditions, such as the initial position and initial velocity. This would help you determine the constants in the equation of motion.
  • Maximum or minimum position: You might be asked to find the maximum or minimum position of the object. This would involve finding the vertex of the parabola in the position vs. time graph.
  • Time intervals: You might be asked to find the time interval during which the object is moving in a certain direction or within a certain range of positions.
  • Different equations of motion: The equation of motion might be different. For example, it could be a cubic function, a trigonometric function, or even a piecewise function.

By understanding the fundamental principles and practicing different types of problems, you'll be well-equipped to tackle any motion problem that comes your way.

Real-World Applications

Kinematics isn't just a theoretical concept; it has numerous real-world applications. Here are a few examples:

  • Sports: Analyzing the motion of a ball in baseball or the trajectory of a basketball shot.
  • Engineering: Designing the suspension system of a car or the flight path of a rocket.
  • Animation: Creating realistic movement for characters in video games and movies.
  • Traffic planning: Optimizing traffic flow and designing safe intersections.

Understanding kinematics allows us to make predictions and control the motion of objects in various situations. It's a fundamental part of physics and engineering.

Conclusion (Part 2)

We've covered a lot of ground in this discussion! We started with a classic motion problem, solved it step-by-step, and then delved deeper into the concepts of kinematics. We explored velocity, acceleration, graphical representations, variations of the problem, and real-world applications. Remember, the key to mastering these concepts is practice and a willingness to think critically about the results you obtain.

Keep exploring, keep questioning, and keep solving! Physics is a fascinating subject that can help you understand the world around you in a whole new way. And if you ever get stuck, don't hesitate to ask for help. There's a whole community of people out there who are passionate about physics and eager to share their knowledge. Now go out there and conquer those motion problems, guys! You've got this!