Motorcycle Acceleration Problem Calculation Of Overtaking Point

In this article, we will delve into a classic physics problem involving a motorcycle accelerating from a standstill and a car moving at a constant velocity. This scenario allows us to explore the concepts of kinematics, including acceleration, velocity, and displacement. We will analyze the motion of both vehicles to determine the point at which the motorcycle overtakes the car. This is a quintessential problem often encountered in introductory physics courses and serves as a great example of how to apply kinematic equations to real-world situations. Understanding these principles is crucial for anyone delving into the world of physics, as they form the bedrock for more advanced concepts in mechanics and dynamics. Let's explore how we can break down this problem, apply the relevant equations, and arrive at a comprehensive solution. This exploration will not only solidify your understanding of kinematic principles but also enhance your problem-solving skills in physics.

Imagine a motorcycle sitting at a red light. The moment the light turns green, the motorcycle accelerates at a constant rate of 4.2 m/s². Simultaneously, a car passes the motorcycle, maintaining a constant velocity of 72 km/h. Our goal is to analyze this scenario and determine when and where the motorcycle will overtake the car. This problem provides a practical application of kinematic equations and requires us to consider the motion of two objects simultaneously. To solve this problem effectively, we need to convert units to ensure consistency, apply the relevant equations of motion for constant acceleration and constant velocity, and then analyze the results. The problem highlights the difference between accelerated motion and uniform motion, and understanding this difference is key to solving it. By working through this problem, we can gain a deeper understanding of how these concepts interact in a real-world scenario.

Before we can begin our calculations, it's essential to ensure that all our units are consistent. The car's velocity is given in kilometers per hour (km/h), while the motorcycle's acceleration is in meters per second squared (m/s²). To avoid confusion and errors, we need to convert the car's velocity from km/h to m/s. This conversion is a fundamental step in many physics problems, as using inconsistent units can lead to incorrect results. To convert km/h to m/s, we multiply by a conversion factor of 1000 meters/kilometer and divide by 3600 seconds/hour. This conversion factor is derived from the fact that there are 1000 meters in a kilometer and 3600 seconds in an hour. Once we've performed this conversion, all our quantities will be in SI units (meters and seconds), making our calculations much smoother and more reliable. This initial step is a crucial foundation for the rest of the problem-solving process.

To convert 72 km/h to m/s, we use the following conversion:

72 km/h * (1000 m/km) / (3600 s/h) = 20 m/s

Now that we have our units aligned, we can turn to the kinematic equations that will help us describe the motion of the motorcycle and the car. Kinematic equations are a set of equations that describe the motion of objects with constant acceleration. These equations relate displacement, initial velocity, final velocity, acceleration, and time. They are a cornerstone of classical mechanics and are essential for analyzing a wide range of motion problems. For this problem, we will use two primary equations: one for the displacement of the motorcycle under constant acceleration and another for the displacement of the car moving at constant velocity. By applying these equations, we can create mathematical models of the motion of each vehicle and then use these models to answer our questions about when and where the motorcycle overtakes the car. Understanding these equations and how to apply them is a fundamental skill in physics. The first equation we'll use describes the displacement of an object under constant acceleration:

x = x₀ + v₀t + (1/2)at²

where:

• x is the final position
• x₀ is the initial position
• v₀ is the initial velocity
• a is the acceleration
• t is the time

Since the motorcycle starts from rest (v₀ = 0) at the origin (x₀ = 0), the equation simplifies to:

x = (1/2)at²

For the car moving at constant velocity, the equation is:

x = x₀ + vt

Since the car passes the motorcycle at the origin (x₀ = 0), the equation simplifies to:

x = vt

The key to finding when the motorcycle overtakes the car lies in recognizing that at the point of overtaking, both vehicles will have the same displacement (x). This means that the final position of the motorcycle will be equal to the final position of the car at the time of overtaking. This concept allows us to set the two displacement equations equal to each other and solve for the time (t) at which this occurs. This is a common strategy in physics problems involving the intersection of motion paths. By equating the two equations, we create a single equation with one unknown variable (t), which we can then solve using algebraic techniques. The solution for t will give us the time it takes for the motorcycle to overtake the car. Once we have the time, we can then substitute it back into either equation to find the position (x) at which the overtaking occurs. This approach demonstrates how mathematical modeling can be used to solve real-world physics problems.

To find the overtaking point, we set the motorcycle's displacement equal to the car's displacement:

(1/2)at² = vt

where:

• a = 4.2 m/s² (motorcycle's acceleration)
• v = 20 m/s (car's velocity)

Now, let's solve the equation for the time (t) when the motorcycle overtakes the car. We have the equation (1/2)at² = vt, and our goal is to isolate t. To do this, we can first rearrange the equation to bring all the terms to one side, resulting in a quadratic equation. Then, we can solve for t using either factoring or the quadratic formula. In this case, factoring is a simpler approach. By factoring out a t from both terms, we can reduce the equation to a simpler form that is easily solved. This process highlights the importance of algebraic manipulation in physics problem-solving. The solution for t will give us the time at which the motorcycle's displacement is equal to the car's displacement, which is the time of overtaking. It's important to note that quadratic equations often have two solutions, but in this physical context, only one solution will be meaningful (the positive one, since time cannot be negative).

Rearranging the equation, we get:

(1/2)at² - vt = 0

Factoring out t:

t((1/2)at - v) = 0

This gives us two possible solutions for t:

1. t = 0 (This corresponds to the initial moment when both vehicles are at the same position.)
2. (1/2)at - v = 0

Solving the second equation for t:

(1/2)at = v
t = (2v) / a
t = (2 * 20 m/s) / 4.2 m/s²
t ≈ 9.52 seconds

With the time (t) at which the motorcycle overtakes the car now calculated, we can determine the distance (x) from the starting point where this overtaking occurs. To do this, we can substitute the value of t back into either the motorcycle's displacement equation or the car's displacement equation. Since both vehicles have the same displacement at the overtaking point, using either equation will yield the same result. This step demonstrates how we can use the time we calculated to find another important quantity in the problem. Substituting the value of t into one of the displacement equations allows us to find the position where the overtaking occurs, providing a complete solution to the problem. This calculation reinforces the connection between time, displacement, velocity, and acceleration in kinematic problems.

We can use the car's equation:

x = vt
x = 20 m/s * 9.52 s
x ≈ 190.4 meters

In conclusion, the motorcycle overtakes the car approximately 9.52 seconds after the light turns green, at a distance of about 190.4 meters from the starting point. This problem has illustrated how we can apply kinematic equations to analyze the motion of objects with constant acceleration and constant velocity. By converting units, applying the appropriate equations, and solving for the unknowns, we were able to determine both the time and the distance at which the motorcycle overtook the car. This type of problem is a fundamental example in physics education, demonstrating the practical application of kinematic principles. The ability to solve these problems is crucial for understanding more complex concepts in mechanics and dynamics. This exercise also highlights the importance of careful unit management and algebraic manipulation in physics problem-solving. The concepts and techniques used in this problem can be applied to a wide range of real-world scenarios, making it a valuable learning experience.