Overview of the Problem
Physics problems often include scenarios like a bus applying brakes and coming to a complete stop. Let's analyze a specific example where a bus is running with a velocity of 60 km/h and then applies brakes, resulting in a retardation (deceleration) of 0.5 m/s2. We aim to determine the distance it will travel before coming to rest. This problem can be approached using basic kinematic equations, specifically focusing on velocity, acceleration, and distance.
Problem Setup and Initial Conditions
Given:
Initial velocity, (u 60) km/h
Final velocity, (v 0) m/s
Deceleration, (a -0.5) m/s2
Step-by-Step Solution
The key equation to use here is (v^2 - u^2 2as), where:
(v) is the final velocity, (u) is the initial velocity, (a) is the acceleration (or deceleration, in this case, hence negative), (s) is the distance covered.Substituting the given values:
[v^2 - u^2 2as]
[0 - left(frac{50}{3}right)^2 2(-0.5)s]
[-frac{2500}{9} -s]
[s frac{2500}{9} 277.8,text{m}]
Therefore, the bus will travel 277.8 meters before coming to rest.
Alternative Approaches and Common Mistakes
Let's look at additional attempts at solving the same problem using different formulas:
Second Method
The second method uses the equation (v^2 - u^2 2as) again, but with a different calculation for initial velocity and acceleration:
[v^2 - u^2 2as]
[0 - (16.67)^2 2(-0.5)s]
[-277.89 -s]
[s 277.89,text{m}]
This result is consistent with the first method.
Third Method
The third method uses the kinematic equations (v u at) and (s ut frac{1}{2}at^2):
Determine acceleration first:
[v u at]
[0 16.67 (-8.34)t]
[-8.34 frac{16.67}{t}]
[t 2,text{seconds}]
Now calculate distance:
[s ut frac{1}{2}at^2]
[s 16.67(2) frac{1}{2}(-8.34)(2)^2]
[s 33.34 - 16.68 16.66,text{m}]
The third method yields a significantly different result, indicating a potential misunderstanding or error in the problem's setup. It's crucial to ensure the units and initial conditions are correctly applied.
Fourth Method
The fourth method examines a different problem with an initial velocity of 72 km/h:
(u 72) km/h 20 m/s
(v 0) m/s
(a -0.1) m/s2
[v^2 - u^2 2as]
[0 - (20)^2 2(-0.1)s]
[-400 -0.2s]
[s frac{400}{0.2} 2000,text{m}]
This method shows that the bus will travel 2 kilometers to come to a stop.
Conclusion
The correct analysis and application of kinematic equations are essential in solving such problems. The first and second methods yield consistent results of 277.8 meters. The third method's result of only 16.7 meters suggests a miscalculation, possibly due to incorrect application of units or acceleration. The fourth method demonstrates how different initial conditions can significantly affect the results.
Understanding and applying these concepts correctly in a real-world setting, such as in road safety or vehicle dynamics, is crucial. If you have further questions or need additional help with similar problems, feel free to ask.