Kinematics 1 I The Chase Answers

Kinematics, the study of motion without considering its causes, forms the bedrock of classical mechanics. Within kinematics lies a subset of problems known as "chase problems," which involve analyzing the relative motion of two or more objects pursuing each other. These scenarios are not only intellectually stimulating but also have practical applications in fields like robotics, game development, and traffic control. Understanding the underlying principles of kinematics and how they apply to chase problems can unlock a deeper understanding of motion and its complexities.

Understanding the Fundamentals

Before diving into the intricacies of chase problems, it's crucial to have a firm grasp of the fundamental concepts of kinematics:

• Displacement: This refers to the change in position of an object. It's a vector quantity, meaning it has both magnitude (the distance traveled) and direction.
• Velocity: This is the rate of change of displacement with respect to time. It's also a vector quantity, indicating both speed and direction.
• Acceleration: This is the rate of change of velocity with respect to time. Like displacement and velocity, it's a vector quantity.

These concepts are interconnected through a set of equations known as the kinematic equations. These equations apply when acceleration is constant and motion is in a straight line:

1. v = u + at (Final velocity = Initial velocity + (Acceleration * Time))
2. s = ut + (1/2)at^2 (Displacement = (Initial velocity * Time) + (1/2 * Acceleration * Time^2))
3. v^2 = u^2 + 2as (Final velocity^2 = Initial velocity^2 + (2 * Acceleration * Displacement))
4. s = (u+v)t/2 (Displacement = ((Initial velocity + Final velocity) * Time) / 2)

Where:

• v = final velocity
• u = initial velocity
• a = acceleration
• t = time
• s = displacement

Understanding how to use these equations is essential for solving a wide range of kinematic problems, including chase scenarios.

The Essence of Chase Problems

Chase problems typically involve two objects, a "pursuer" and an "evader," moving along the same path. The goal is usually to determine:

• When and where the pursuer will catch the evader.
• The minimum speed or acceleration the pursuer needs to catch the evader.
• The closest distance between the pursuer and the evader.

The key to solving these problems lies in analyzing the relative motion of the two objects. This means considering the evader's motion from the pursuer's frame of reference, or vice-versa.

Strategies for Tackling Chase Problems

Here's a step-by-step approach to solving chase problems effectively:

1. Understand the Scenario: Carefully read the problem statement and visualize the scenario. Draw a diagram if necessary, labeling the initial positions, velocities, and accelerations of both the pursuer and the evader.

2. Define Variables: Assign variables to all relevant quantities, such as the initial positions (s₀), initial velocities (u), accelerations (a), and time (t) for both the pursuer and the evader. Be consistent with your notation. For example, you might use subscripts "p" and "e" to denote the pursuer and evader, respectively (e.g., uₚ for the pursuer's initial velocity).

3. Establish a Coordinate System: Choose a convenient coordinate system with a defined origin and direction. This will help you keep track of the positions and displacements of the objects.

4. Write Kinematic Equations: Write down the relevant kinematic equations for both the pursuer and the evader. The equations you use will depend on the information given in the problem statement and what you are trying to find.

5. Relate the Motions: The crucial step is to relate the motions of the pursuer and the evader. This usually involves setting up an equation that expresses the condition for the pursuer to catch the evader. For example, if the pursuer catches the evader at time t, then their positions at time t must be equal:

   sₚ(t) = sₑ(t)

   Where sₚ(t) and sₑ(t) represent the positions of the pursuer and evader, respectively, as functions of time.

6. Solve the Equations: Solve the system of equations you have created. This may involve algebraic manipulation, substitution, or solving quadratic equations.

7. Check Your Answer: Once you have found a solution, check if it makes sense in the context of the problem. For example, the time taken for the pursuer to catch the evader should be positive.

Illustrative Examples

Let's illustrate these strategies with a few examples:

Example 1: The Simple Chase

Problem: A car starts from rest and accelerates at 2 m/s² to catch a truck moving with a constant velocity of 10 m/s. The truck is initially 20 meters ahead of the car. How long does it take the car to catch the truck, and how far does the car travel?

Solution:

1. Understand the Scenario: A car accelerates to catch a truck moving at a constant speed.

2. Define Variables:

   • Car: uₚ = 0 m/s, aₚ = 2 m/s², s₀ₚ = 0 m
   • Truck: uₑ = 10 m/s, aₑ = 0 m/s², s₀ₑ = 20 m

3. Establish a Coordinate System: Let the initial position of the car be the origin (s = 0).

4. Write Kinematic Equations:

   • Car: sₚ(t) = uₚt + (1/2)aₚt² = 0 + (1/2)(2)t² = t²
   • Truck: sₑ(t) = s₀ₑ + uₑt + (1/2)aₑt² = 20 + 10t + 0 = 20 + 10t

5. Relate the Motions: The car catches the truck when their positions are equal:

   t² = 20 + 10t

6. Solve the Equations: Rearrange the equation to form a quadratic:

   t² - 10t - 20 = 0

   Using the quadratic formula:

   t = (10 ± √(10² - 4(1)(-20))) / (2(1)) = (10 ± √(180)) / 2 = (10 ± 6√5) / 2 = 5 ± 3√5

   Since time must be positive, we take the positive root:

   t = 5 + 3√5 ≈ 11.71 seconds

   The distance traveled by the car is:

   sₚ(t) = t² = (11.71)² ≈ 137.12 meters

7. Check Your Answer: The time is positive, and the distance traveled by the car is reasonable.

Example 2: The Minimum Speed Chase

Problem: A police car is parked on the side of the road. A speeding car passes it at a constant speed of 30 m/s. The police car starts accelerating at 4 m/s² the instant the speeding car passes. What is the minimum initial distance the police car must be from the next intersection, which is 500 meters away, to catch the speeding car before it reaches the intersection?

Solution:

1. Understand the Scenario: A police car accelerates to catch a speeding car before it reaches an intersection.

2. Define Variables:

   • Police Car: uₚ = 0 m/s, aₚ = 4 m/s², s₀ₚ = 0 m (relative to the speeding car's initial position)
   • Speeding Car: uₑ = 30 m/s, aₑ = 0 m/s², s₀ₑ = 0 m

3. Establish a Coordinate System: Let the initial position of both cars be the origin (s = 0).

4. Write Kinematic Equations:

   • Police Car: sₚ(t) = (1/2)aₚt² = (1/2)(4)t² = 2t²
   • Speeding Car: sₑ(t) = uₑt = 30t

5. Relate the Motions: The police car catches the speeding car when their positions are equal:

   2t² = 30t

6. Solve the Equations:

   2t² - 30t = 0 2t(t - 15) = 0

   This gives us two solutions: t = 0 (when they are initially together) and t = 15 seconds (when the police car catches the speeding car).

   At t = 15 seconds, the speeding car has traveled:

   sₑ(15) = 30 * 15 = 450 meters

   This means the police car catches the speeding car 450 meters from the starting point.

   Since the intersection is 500 meters away, the police car catches the speeding car 50 meters before the intersection. Therefore, the minimum initial distance the police car must be from the intersection is:

   500 meters - 450 meters = 50 meters

7. Check Your Answer: The time and distance are reasonable, and the police car catches the speeding car before the intersection.

Example 3: The Relative Velocity Approach

Problem: Two cars, A and B, are traveling in the same direction. Car A is traveling at 20 m/s and car B is traveling at 30 m/s. If car B is initially 100 meters behind car A, how long will it take car B to catch car A?

Solution:

1. Understand the Scenario: Car B, moving faster, chases car A.

2. Define Variables:

   • Car A: uₐ = 20 m/s, aₐ = 0 m/s², s₀ₐ = 0 m
   • Car B: uв = 30 m/s, aв = 0 m/s², s₀в = -100 m

3. Establish a Coordinate System: Let the initial position of car A be the origin (s = 0).

4. Write Kinematic Equations:

   • Car A: sₐ(t) = uₐt = 20t
   • Car B: sв(t) = s₀в + uвt = -100 + 30t

5. Relate the Motions: The car B catches the car A when their positions are equal:

   20t = -100 + 30t

6. Solve the Equations:

   10t = 100 t = 10 seconds

7. Check Your Answer: The time is positive and reasonable.

Alternative Solution using Relative Velocity:

The relative velocity of car B with respect to car A is v_rel = vв - vₐ = 30 m/s - 20 m/s = 10 m/s.

The initial distance between them is 100 meters. The time it takes for car B to close this distance is:

t = distance / relative velocity = 100 m / 10 m/s = 10 seconds

This confirms the previous solution.

Advanced Considerations

While the examples above illustrate the basic principles, some chase problems can be more complex, involving:

• Non-Constant Acceleration: If the acceleration of either the pursuer or the evader is not constant, the kinematic equations cannot be directly applied. Instead, you'll need to use calculus to determine the velocity and position as functions of time.
• Two-Dimensional Motion: If the motion is not confined to a straight line, you'll need to analyze the motion in two dimensions (e.g., x and y coordinates) separately. This will involve resolving velocities and accelerations into their components.
• Constraints: Some problems may involve constraints, such as a maximum speed or acceleration that the pursuer can achieve. These constraints will need to be taken into account when solving the problem.

Practical Applications

Chase problems are not just theoretical exercises; they have numerous practical applications:

• Robotics: Robots often need to track and intercept moving objects. Chase problem techniques can be used to design algorithms for robot navigation and control.
• Game Development: Chase scenarios are common in video games, where players control characters that chase or evade other characters. Understanding the physics of chase problems can help create realistic and engaging gameplay.
• Traffic Control: Analyzing the motion of vehicles in traffic can be modeled as a chase problem. This can be used to develop strategies for preventing collisions and improving traffic flow.
• Military Applications: Interception of missiles or aircraft utilizes principles of chase problems, considering variables like speed, trajectory, and acceleration to predict and achieve successful interception.
• Sports: Many sports, like soccer or basketball, involve players chasing after a ball or other players. While not explicitly calculated, the players intuitively use principles related to chase problems to intercept or evade opponents.

Tips for Success

• Practice Regularly: The best way to master chase problems is to practice solving a variety of examples.
• Draw Diagrams: Visualizing the scenario can help you understand the problem and identify the relevant variables.
• Be Organized: Keep your work neat and organized to avoid making mistakes.
• Check Your Units: Make sure that all your units are consistent.
• Think Critically: Don't just blindly apply formulas. Think about the physical meaning of the equations and the assumptions you are making.
• Use Technology: Utilize graphing calculators or computer software to solve complex equations or visualize the motion.

The Interplay of Kinematics and Calculus

As motion scenarios become more intricate, involving non-constant acceleration, the basic kinematic equations are no longer sufficient. This is where calculus becomes an indispensable tool.

Calculus in Kinematics:

• Velocity as a Derivative: Instantaneous velocity v(t) is the derivative of the position function s(t) with respect to time: v(t) = ds(t)/dt.
• Acceleration as a Derivative: Instantaneous acceleration a(t) is the derivative of the velocity function v(t) with respect to time: a(t) = dv(t)/dt. It's also the second derivative of the position function: a(t) = d²s(t)/dt².
• Position and Velocity as Integrals: If acceleration a(t) is known, velocity v(t) can be found by integrating a(t) with respect to time: v(t) = ∫ a(t) dt. Similarly, position s(t) can be found by integrating v(t) with respect to time: s(t) = ∫ v(t) dt.

Chase Problems with Calculus:

When acceleration is a function of time a(t), you must integrate to find velocity and position. For example, suppose a pursuer has an acceleration a(t) = kt, where k is a constant. To find its velocity:

v(t) = ∫ kt dt = (1/2)kt² + C₁

Where C₁ is the constant of integration, representing the initial velocity v₀. So, v(t) = (1/2)kt² + v₀.

Similarly, to find the position:

s(t) = ∫ [(1/2)kt² + v₀] dt = (1/6)kt³ + v₀t + C₂

Where C₂ is another constant of integration, representing the initial position s₀. Thus, s(t) = (1/6)kt³ + v₀t + s₀.

These expressions for velocity and position can then be used to set up and solve chase problems, similar to the constant acceleration case, but involving more complex algebraic manipulations.

Example:

A car's acceleration is given by a(t) = 2t m/s². It starts from rest at the origin. A truck moves at a constant speed of 10 m/s and is initially 20 meters ahead. Find when and where the car catches the truck.

1. Car's Velocity: v(t) = ∫ 2t dt = t² + C₁. Since it starts from rest, v(0) = 0, so C₁ = 0 and v(t) = t².
2. Car's Position: s(t) = ∫ t² dt = (1/3)t³ + C₂. Since it starts at the origin, s(0) = 0, so C₂ = 0 and s(t) = (1/3)t³.
3. Truck's Position: s_truck(t) = 20 + 10t.
4. Equate Positions: (1/3)t³ = 20 + 10t.

Solving this cubic equation for t is more complex but can be done numerically or graphically to find the time when the car catches the truck. Then, substitute t back into either position equation to find the location.

Conclusion

Kinematics, particularly chase problems, offers a fascinating glimpse into the world of motion. By understanding the fundamental concepts, mastering the problem-solving strategies, and practicing regularly, you can unlock a deeper understanding of how objects move and interact with each other. Moreover, the connection with calculus extends these concepts to more complex scenarios. Whether you're a student, engineer, or simply someone curious about the world around you, the principles of kinematics and chase problems provide a valuable framework for analyzing and understanding the dynamics of motion.