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In a historical movie, two knights on horseback start from rest \(88.0 \mathrm{m}\) apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of \(0.300 \mathrm{m} / \mathrm{s}^{2},\) while Sir Alfred's has a magnitude of \(0.200 \mathrm{m} / \mathrm{s}^{2} .\) Relative to Sir George's starting point, where do the knights collide?

Short Answer

Expert verified
The knights collide 52.8 meters from Sir George's starting point.

Step by step solution

01

Understanding the Problem

We have two knights starting 88 meters apart. Sir George accelerates at 0.300 \( \text{m/s}^2 \), and Sir Alfred accelerates at 0.200 \( \text{m/s}^2 \). We need to find at what position the knights collide relative to Sir George's starting point.
02

Setting up the Equations of Motion

Using the equation of motion \(d = \frac{1}{2} a t^2\), we can express the distance each knight travels. Let \(d_G\) and \(d_A\) be the distances traveled by Sir George and Sir Alfred, respectively. Then: \[d_G = \frac{1}{2} \times 0.300 \times t^2\] \[d_A = \frac{1}{2} \times 0.200 \times t^2\] The total distance traveled by both when they collide is 88 meters: \[d_G + d_A = 88\]
03

Substitute and Combine Equations

Substitute the expressions for \(d_G\) and \(d_A\) into the total distance equation:\[\frac{1}{2} \times 0.300 t^2 + \frac{1}{2} \times 0.200 t^2 = 88\] Combine and simplify:\[0.150t^2 + 0.100t^2 = 88\] \[0.250t^2 = 88\]
04

Solving for Time

Solve for \(t^2\) by dividing both sides of the equation by 0.250:\[t^2 = \frac{88}{0.250} = 352\] Then, take the square root of both sides to find \(t\): \[t = \sqrt{352} \approx 18.76 \text{ seconds}\]
05

Finding the Collision Point

Use \(t = 18.76\) seconds to find \(d_G\), the distance Sir George travels:\[d_G = \frac{1}{2} \times 0.300 \times (18.76)^2\] Calculating:\[d_G = 0.150 \times 352 = 52.8 \text{ meters}\]
06

Conclusion

The knights will collide 52.8 meters from Sir George's starting point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equations of Motion
Equations of motion are fundamental to solving problems in kinematics. When dealing with objects in motion like the knights in the exercise, we commonly use the equation \( d = \frac{1}{2} at^2 \). This formula helps us calculate the distance \( d \) that an object travels over time \( t \) under constant acceleration \( a \).

In our knight exercise, this equation allows us to determine the distances traveled by Sir George and Sir Alfred as they accelerate toward each other. By setting up individual equations for both, we ensure a structured approach to understanding how each knight moves. This not only helps in finding when they meet but also establishes a foundation for more complex motion problems.

Always start by identifying the right equation based on the given parameters. Is the object starting from rest, or does it have an initial velocity? These factors guide which formula to use. Equations of motion simplify many real-world problems by providing a clear mathematical path to the solution.

Remember: Always list what you know and substitute wisely into the equations.
Acceleration
Acceleration refers to the rate at which an object changes its velocity. It's a vector quantity, which means it has both magnitude and direction. When the knights begin their ride, they start with zero velocity and accelerate toward each other at 0.300 \( \text{m/s}^2 \) and 0.200 \( \text{m/s}^2 \) for Sir George and Sir Alfred respectively.

Understanding acceleration is crucial because it defines how quickly the knights pick up speed as they close the gap between them. It's calculated by taking the change in velocity over a specific time period, displayed as \( a = \frac{\Delta v}{t} \).

In practical terms, acceleration impacts how swiftly an object reaches a certain speed. In this scenario, Sir George accelerates more quickly than Sir Alfred, meaning he covers more distance in the same time span. This difference affects where and when they collide.

Think about acceleration as the 'push' that causes stationary objects to move and moving objects to stop or change their velocity.
Relative Motion
Relative motion describes how two objects move concerning each other. In our exercise, the two knights are not just moving independently but directly toward each other. This concept of relative motion is key to understanding how their paths intersect.

When analyzing relative motion, it's beneficial to focus on a reference point. Here, Sir George's starting position serves as such. From this point, we measure how far each knight travels before they meet. This tactic simplifies the math by visualizing one object's movement in relation to another's.

By considering their combined movement, it becomes easier to figure out their meeting point. We know that the total distance is 88 meters. Hence, the goal is to find how much of that distance each travels before they cross paths.

Use relative motion concepts when objects are in direct pathways or share a common destination but start from different positions.
Distance Calculation
Distance calculation is essential in determining how far objects travel, especially when they start from rest and under acceleration. For our knights, the key is to calculate how much ground each covers using their respective acceleration values and the time it takes for them to collide.

To achieve precise distance measurements, we used the formula \( d = \frac{1}{2} at^2 \) for both knights. This approach, combined with the total clash distance, allowed us to determine how far from the starting point of Sir George the two meet.

Sir George, accelerating at 0.300 \( \text{m/s}^2 \), covers more distance in the same time frame than Sir Alfred, helping define the collision point. Simple arithmetic, such as addition and subtraction, often completes these problems once distances are calculated.

These calculations offer insights into real-world motion scenarios, ensuring accuracy when predicting object locations at certain times.

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