Bus Motion Problem: Acceleration, Speed & Distance Solved
Hey everyone! Let's dive into a classic physics problem involving a bus, acceleration, speed, and distance. We'll break down the scenario step-by-step to make sure we understand the concepts and calculations involved. So, buckle up, and let's get started!
Understanding the Problem Scenario
Our problem describes a bus that undergoes three distinct phases of motion:
- Acceleration: The bus starts from rest and accelerates at a constant rate. This is where the bus picks up speed. We're given the acceleration rate and the time duration for this phase.
- Constant Speed: After accelerating, the bus maintains a constant speed for a certain period. This means the bus is neither speeding up nor slowing down during this phase.
- Deceleration (Braking): Finally, the brakes are applied, causing the bus to decelerate (or experience negative acceleration) until it comes to a complete stop. We're given the time it takes to stop during this phase.
Our goal is to determine three key things:
- The constant speed the bus maintains during the second phase.
- The acceleration (or deceleration) during the braking phase.
- The total distance traveled by the bus throughout the entire journey.
To solve this, we'll need to use our knowledge of kinematics, which is the branch of physics that deals with the motion of objects. The key equations we'll be using are those that relate displacement, velocity, acceleration, and time under constant acceleration. These are often referred to as the equations of motion or SUVAT equations (where SUVAT stands for Displacement, Initial Velocity, Final Velocity, Acceleration, and Time).
Phase 1: Acceleration from Rest
In this initial phase, the bus starts from rest, which means its initial velocity (v₀) is 0 m/s. The bus then accelerates at a rate (a) of 1.8 m/s² for a time (t) of 12 seconds. To figure out the bus's final velocity (v) at the end of this phase, we can use one of the fundamental equations of motion:
v = vâ‚€ + at
Plugging in the values we have:
v = 0 + (1.8 m/s²)(12 s) = 21.6 m/s
So, at the end of the acceleration phase, the bus is traveling at a speed of 21.6 m/s. This speed becomes crucial for understanding the next phase of the journey. In this phase, it's also important to calculate the distance traveled (s), which we can find using another equation of motion:
s = v₀t + (1/2)at²
Substituting our values:
s = (0 m/s)(12 s) + (1/2)(1.8 m/s²)(12 s)² = 129.6 m
Therefore, during the acceleration phase, the bus covers a distance of 129.6 meters. This information, alongside the final velocity, sets the stage for analyzing the subsequent phases of the bus's journey.
Phase 2: Constant Speed
After the acceleration phase, the bus cruises at a constant speed for 5 seconds. Remember, the final velocity from the previous phase (21.6 m/s) is now the constant speed during this phase. Since the speed is constant, there's no acceleration (a = 0 m/s²). This makes calculations a bit simpler. To find the distance traveled during this phase (s), we can use the basic formula:
***s = vt***
Where v is the constant speed and t is the time. Plugging in our values:
***s = (21.6 m/s)(5 s) = 108 m***
So, during the 5 seconds of constant speed, the bus covers a distance of 108 meters. Understanding this phase is crucial as it provides a transition between the acceleration and deceleration periods, adding another layer to our understanding of the bus's complete journey.
Phase 3: Deceleration (Braking)
Now, things get a bit more interesting. The bus driver applies the brakes, and the bus decelerates to a stop in 10 seconds. The initial velocity (vâ‚€) for this phase is the constant speed we calculated earlier, 21.6 m/s. The final velocity (v) is 0 m/s since the bus comes to a stop. We know the time (t) is 10 seconds. First, we need to calculate the acceleration (a) during braking, which will be a negative value because it's deceleration. We can use the equation:
***v = vâ‚€ + at***
Rearranging for a:
***a = (v - vâ‚€) / t***
Substituting the values:
***a = (0 m/s - 21.6 m/s) / 10 s = -2.16 m/s²***
The acceleration during braking is -2.16 m/s², which means the bus is slowing down at this rate. Next, we need to find the distance traveled during braking. We can use the equation:
***s = v₀t + (1/2)at²***
Plugging in the values:
***s = (21.6 m/s)(10 s) + (1/2)(-2.16 m/s²)(10 s)² = 108 m***
So, the bus travels 108 meters while braking. This final phase is key to understanding the total distance traveled and the dynamics of how the bus slows down to a halt.
Calculating the Total Distance
To find the total distance traveled by the bus, we simply add up the distances from each phase:
- Phase 1 (Acceleration): 129.6 meters
- Phase 2 (Constant Speed): 108 meters
- Phase 3 (Braking): 108 meters
Total Distance = 129.6 m + 108 m + 108 m = 345.6 meters
Therefore, the bus travels a total distance of 345.6 meters during its journey. This total distance gives us a comprehensive view of the bus's entire motion, combining the acceleration, constant speed, and deceleration phases.
Summary of Results
Let's recap our findings:
- Constant Speed: The bus maintains a constant speed of 21.6 m/s during the second phase.
- Acceleration During Braking: The acceleration during braking is -2.16 m/s².
- Total Distance Traveled: The bus travels a total distance of 345.6 meters.
These results provide a complete picture of the bus's motion, from its initial acceleration to its final stop. Understanding each phase and how they connect is crucial in grasping the full dynamics of the problem.
Conclusion: Kinematics in Action
So, there you have it! We've successfully solved a multi-stage motion problem involving a bus, applying the principles of kinematics. By breaking down the problem into phases and using the appropriate equations of motion, we were able to determine the constant speed, the acceleration during braking, and the total distance traveled. Problems like these are a great way to see physics in action and to sharpen our problem-solving skills. Keep practicing, and you'll become a motion master in no time! Remember, physics is all around us, and understanding these concepts helps us make sense of the world in a whole new way.
