Finding Maximum Speed Uniform Acceleration Constant Speed And Uniform Deceleration
Introduction
In physics, understanding the motion of objects is a fundamental concept. Kinematics, the study of motion without considering its causes, provides the tools to analyze scenarios involving displacement, velocity, and acceleration. This article delves into a classic problem involving a car undergoing three distinct phases of motion: uniform acceleration, constant speed, and uniform deceleration. By applying kinematic principles and utilizing the given information about the total distance traveled and the time intervals for each phase, we aim to determine the maximum speed attained by the car. This exploration not only reinforces our understanding of motion but also highlights the practical application of physics in everyday scenarios. Let's embark on a journey to unravel the dynamics of this car's movement and discover its peak velocity.
Problem Statement
A car begins its journey by accelerating uniformly for a period of 4 seconds. Following this acceleration phase, the car maintains a constant speed for 10 seconds. Finally, the car decelerates uniformly, coming to a complete stop in 6 seconds. The total distance covered during this entire motion is 600 meters. Our objective is to calculate the maximum speed achieved by the car during its journey. This problem requires us to dissect the motion into three distinct phases and apply appropriate kinematic equations to each phase. By carefully analyzing the relationships between acceleration, velocity, time, and displacement, we can determine the peak velocity attained by the car. The solution involves a blend of conceptual understanding and mathematical application, showcasing the power of physics in solving real-world problems. Let's dive into the details and uncover the solution step by step.
Solution Approach
To solve this problem, we will break down the car's motion into three distinct phases: uniform acceleration, constant speed, and uniform deceleration. For each phase, we will apply the appropriate kinematic equations to relate displacement, initial velocity, final velocity, acceleration, and time. We will use the information provided about the time intervals and the total distance traveled to form a system of equations. By solving this system, we can determine the maximum speed of the car, which occurs during the constant speed phase. Let's define the variables and equations for each phase:
- Phase 1: Uniform Acceleration (0 to 4 seconds)
- Initial velocity: v₀ = 0 m/s (assuming the car starts from rest)
- Final velocity: v₁ (this is the maximum speed we want to find)
- Time: t₁ = 4 s
- Acceleration: a₁
- Displacement: s₁
- Phase 2: Constant Speed (4 to 14 seconds)
- Velocity: v₂ = v₁ (constant speed)
- Time: t₂ = 10 s
- Acceleration: a₂ = 0 m/s²
- Displacement: s₂
- Phase 3: Uniform Deceleration (14 to 20 seconds)
- Initial velocity: v₃ = v₁
- Final velocity: v₄ = 0 m/s (car comes to rest)
- Time: t₃ = 6 s
- Acceleration: a₃
- Displacement: s₃
Now, let's apply the kinematic equations to each phase.
Phase 1: Uniform Acceleration
During the uniform acceleration phase, the car's velocity increases steadily from rest to its maximum speed. We can use the following kinematic equations:
- v₁ = v₀ + a₁t₁
- s₁ = v₀t₁ + (1/2)a₁t₁²
Since v₀ = 0, these equations simplify to:
- v₁ = a₁t₁ = 4a₁
- s₁ = (1/2)a₁t₁² = 8a₁
These equations relate the maximum speed v₁, the acceleration a₁, and the displacement s₁ during the first phase.
Phase 2: Constant Speed
In this phase, the car maintains a constant speed, which is the maximum speed it achieved during the acceleration phase. The equations for this phase are straightforward:
- v₂ = v₁ (constant speed)
- s₂ = v₂t₂ = v₁t₂ = 10v₁
The displacement s₂ during this phase is simply the product of the maximum speed v₁ and the time t₂.
Phase 3: Uniform Deceleration
During the deceleration phase, the car slows down uniformly from its maximum speed to rest. We can use the following kinematic equations:
- v₄ = v₃ + a₃t₃
- s₃ = v₃t₃ + (1/2)a₃t₃²
Since v₄ = 0 and v₃ = v₁, these equations become:
- 0 = v₁ + 6a₃
- s₃ = 6v₁ + 18a₃
From the first equation, we can express a₃ in terms of v₁: a₃ = -v₁/6. Substituting this into the second equation, we get:
s₃ = 6v₁ + 18(-v₁/6) = 6v₁ - 3v₁ = 3v₁
These equations relate the maximum speed v₁, the deceleration a₃, and the displacement s₃ during the final phase.
Total Distance
The total distance traveled is the sum of the distances traveled in each phase:
s₁ + s₂ + s₃ = 600 m
Substituting the expressions for s₁, s₂, and s₃ in terms of v₁ and a₁, we get:
8a₁ + 10v₁ + 3v₁ = 600
We also have the equation v₁ = 4a₁ from Phase 1. We can substitute a₁ = v₁/4 into the total distance equation:
8(v₁/4) + 10v₁ + 3v₁ = 600 2v₁ + 10v₁ + 3v₁ = 600 15v₁ = 600
Solving for v₁, we find the maximum speed:
v₁ = 600 / 15 = 40 m/s
Answer
Therefore, the maximum speed of the car is 40 m/s. The correct answer is (a) 40 m/s.
Conclusion
In this article, we successfully determined the maximum speed of a car undergoing uniform acceleration, constant speed, and uniform deceleration. By carefully dissecting the motion into three distinct phases and applying the appropriate kinematic equations, we were able to establish relationships between displacement, velocity, acceleration, and time. We then utilized the given information about the total distance traveled and the time intervals to form a system of equations, which we solved to find the peak velocity. This problem exemplifies the power of physics in analyzing real-world scenarios and reinforces the importance of understanding kinematic principles. The solution involved a blend of conceptual understanding and mathematical application, highlighting the elegance and practicality of physics. We hope this exploration has deepened your understanding of motion and its intricacies. Remember, the key to solving such problems lies in breaking them down into manageable parts and applying the relevant principles systematically.
