Determining Maximum Speed Car Uniform Acceleration Constant Speed Deceleration
Understanding the motion of objects is a fundamental aspect of physics, and this article delves into a classic problem involving a car's motion with varying phases: uniform acceleration, constant speed, and uniform deceleration. We will explore how to determine the maximum speed achieved by the car given the time intervals for each phase and the total distance traveled. This problem showcases the application of kinematic equations and the relationship between displacement, velocity, and time.
Problem Statement
A car accelerates uniformly for 4 seconds, then maintains a constant speed for 10 seconds, and finally decelerates uniformly to rest in 6 seconds. If the total distance traveled is 600 meters, what is the maximum speed attained by the car?
(a) 40 m/s (b) -40 m/s
Solution
To solve this problem, we need to break it down into three distinct phases of motion: acceleration, constant speed, and deceleration. We will use the equations of motion for constant acceleration to analyze each phase and then combine the information to find the maximum speed.
Phase 1: Uniform Acceleration
In this initial phase, the car accelerates uniformly for 4 seconds. Let's denote the initial velocity as v₀ (which we'll assume is 0 m/s since the car starts from rest), the final velocity (which is also the maximum speed) as v, the acceleration as a₁, and the time as t₁ = 4 s. The distance traveled during this phase, s₁, can be calculated using the following kinematic equation:
s₁ = v₀ t₁ + (1/2) a₁ t₁²
Since v₀ = 0, the equation simplifies to:
s₁ = (1/2) a₁ t₁² = (1/2) a₁ (4)² = 8 a₁
We also know the relationship between final velocity, initial velocity, acceleration, and time:
v = v₀ + a₁ t₁ = 0 + a₁ (4) = 4 a₁
From this, we can express a₁ in terms of v: a₁ = v/4
Substituting this back into the equation for s₁, we get:
s₁ = 8 * (v/4) = 2 v
Therefore, the distance traveled during the acceleration phase is s₁ = 2v. This equation highlights the direct relationship between the distance covered during uniform acceleration and the final velocity achieved. The longer the distance, the higher the final velocity, assuming constant acceleration.
Understanding the physics behind this phase is crucial. The uniform acceleration means the car's velocity increases at a constant rate. This constant rate of change in velocity allows us to use simplified kinematic equations. The equation s₁ = 2v is a direct result of this constant acceleration and provides a key link between the distance traveled and the maximum speed attained.
The derivation of s₁ = 2v also underscores the power of using kinematic equations to analyze motion. By carefully identifying the known variables (initial velocity, time) and the unknowns (final velocity, distance), we can select the appropriate equations and solve for the desired quantities. This systematic approach is essential for tackling more complex problems in mechanics. Moreover, the ability to express one variable in terms of another (e.g., a₁ in terms of v) is a valuable problem-solving technique. It allows us to reduce the number of unknowns and simplify the equations, leading to a more manageable solution process. This approach not only provides the correct answer but also enhances our understanding of the underlying physical principles.
Phase 2: Constant Speed
During this phase, the car maintains a constant speed v for 10 seconds (t₂ = 10 s). The distance traveled, s₂, is simply the product of speed and time:
s₂ = v t₂ = v (10) = 10 v
The distance traveled at constant speed is directly proportional to the speed and the time interval. This is a straightforward application of the definition of speed: distance traveled per unit time. In this phase, there is no acceleration, meaning the velocity remains unchanged. The simplicity of this phase provides a crucial link between the acceleration and deceleration phases, as the maximum speed attained during acceleration is maintained throughout this constant speed phase.
The equation s₂ = 10v clearly illustrates this relationship. The longer the time spent at constant speed, the greater the distance covered. This phase can be visualized as a rectangle on a velocity-time graph, where the area under the graph (which represents the distance) is the product of the constant velocity and the time interval. Understanding this graphical representation can provide a deeper insight into the motion of the car.
Furthermore, the constant speed phase highlights the concept of inertia. Once the car has reached the maximum speed v, it continues to move at that speed unless acted upon by an external force (such as the brakes in the deceleration phase). This principle is fundamental to Newton's laws of motion. The car's inertia, its resistance to changes in its state of motion, is what allows it to maintain a constant speed in the absence of any net force.
This phase also serves as a practical example of how machines often operate. Vehicles, for instance, often spend a significant portion of their journey traveling at a constant speed, making this phase a relevant scenario for real-world applications. The constant speed phase simplifies the calculations and helps to build a clear picture of the car's overall motion.
Phase 3: Uniform Deceleration
In the final phase, the car decelerates uniformly to rest in 6 seconds (t₃ = 6 s). The initial velocity for this phase is v (the maximum speed), the final velocity is 0 m/s, and the deceleration is a₃. The distance traveled, s₃, can be calculated using the following kinematic equation:
s₃ = v t₃ + (1/2) a₃ t₃²
We also have the relationship:
0 = v + a₃ t₃ = v + a₃ (6)
From this, we can express a₃ in terms of v: a₃ = -v/6
Substituting this back into the equation for s₃, we get:
s₃ = v (6) + (1/2) (-v/6) (6)² = 6v - 3v = 3v
Therefore, the distance traveled during the deceleration phase is s₃ = 3v. This equation reveals the relationship between the distance covered during uniform deceleration and the initial velocity (maximum speed). A higher initial velocity will result in a longer distance covered before coming to a complete stop, assuming uniform deceleration.
The negative sign in a₃ = -v/6 signifies that the car is decelerating, meaning its velocity is decreasing over time. This deceleration is a uniform process, which allows us to apply kinematic equations. The distance s₃ is directly proportional to the initial velocity v, but the constant of proportionality (3) is different from the acceleration phase (2) due to the differing time intervals and the fact that the car is slowing down instead of speeding up.
The physics behind this deceleration phase is crucial in understanding braking systems in vehicles. The deceleration rate is determined by the braking force applied, and the distance required to stop is influenced by both the initial velocity and the deceleration rate. This principle is fundamental to road safety and vehicle design. Understanding the relationship between initial velocity, deceleration, and stopping distance is essential for safe driving practices.
The calculation of s₃ = 3v also demonstrates the importance of choosing the appropriate kinematic equations for the given situation. By using the equations that relate distance, initial velocity, final velocity, acceleration, and time, we can accurately analyze the motion and determine the distance traveled during deceleration. This systematic approach is a key skill in physics problem-solving.
Total Distance and Maximum Speed
The total distance traveled is the sum of the distances from each phase:
s = s₁ + s₂ + s₃ = 2v + 10v + 3v = 15v
We are given that the total distance is 600 meters:
15v = 600
Solving for v, we get:
v = 600 / 15 = 40 m/s
Therefore, the maximum speed attained by the car is 40 m/s.
This final calculation ties together the results from each phase to arrive at the solution. The total distance is the sum of the distances covered during acceleration, constant speed, and deceleration. By setting this total distance equal to the given value (600 meters), we can solve for the unknown maximum speed v. This step highlights the importance of combining the individual components of a problem to arrive at the overall solution.
The result, v = 40 m/s, is the maximum speed the car attained during its motion. This value is a crucial parameter in understanding the car's motion profile. It represents the peak velocity reached during the acceleration phase and maintained during the constant speed phase. This maximum speed directly influences the distances covered in each phase, as well as the overall time taken for the journey.
The entire problem-solving process, from breaking down the motion into phases to combining the results, exemplifies the power of analytical thinking in physics. By carefully considering the different phases of motion and applying the appropriate kinematic equations, we can solve for unknown quantities and gain a comprehensive understanding of the object's movement. This systematic approach is not only applicable to this specific problem but also to a wide range of motion-related scenarios in physics and engineering.
Conclusion
By analyzing the motion of the car in three distinct phases – uniform acceleration, constant speed, and uniform deceleration – we were able to determine the maximum speed attained. The solution, 40 m/s, was found by applying kinematic equations to each phase and then combining the results based on the total distance traveled. This problem demonstrates the importance of understanding the concepts of uniform motion and the ability to apply relevant equations to solve for unknown quantities. The correct answer is (a) 40 m/s.
This problem serves as a valuable example of how to analyze complex motion by breaking it down into simpler components. The key is to identify the distinct phases of motion, apply the appropriate kinematic equations to each phase, and then combine the information to solve for the desired quantity. This approach is applicable to a wide range of physics problems involving motion, and it highlights the power of analytical thinking and problem-solving skills in understanding the physical world.
Furthermore, the problem underscores the significance of the concepts of acceleration, constant speed, and deceleration in real-world scenarios. Understanding these concepts is crucial for analyzing the motion of vehicles, projectiles, and other objects. The kinematic equations provide a powerful tool for predicting and understanding motion, and this problem serves as a practical demonstration of their application.
