Calculating Catch-Up Time: Solving for T in a Bus Chase Scenario

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Discussion Overview

The discussion revolves around a problem involving a man chasing a bus that accelerates away from him. Participants explore the conditions under which the man can catch the bus, the mathematical formulation of the problem, and the implications of the solutions derived from the equations of motion.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant describes the scenario and poses the question of when the man catches the bus, noting that there are two possible times (T) and asks what they represent.
  • Another participant introduces a condition that must be met for the problem to make sense, stating that the man's speed must satisfy a specific inequality related to the initial distance and acceleration of the bus.
  • A third participant provides a mathematical formulation for the distance between the man and the bus as a function of time, leading to a quadratic equation that yields two potential solutions for time.
  • It is noted that if the man's speed squared is equal to twice the initial distance, there is a unique solution, while if it is greater, there are two solutions indicating different moments of interaction between the man and the bus.
  • Some participants engage in a light-hearted debate about the nature of the problem, with one suggesting it is too simple for a brain teaser and others defending the clarity of the explanation provided.

Areas of Agreement / Disagreement

Participants express differing views on the complexity of the problem, with some considering it straightforward while others argue it is more nuanced. There is no consensus on the appropriateness of the problem as a brain teaser, and the discussion includes both mathematical reasoning and informal banter.

Contextual Notes

The discussion includes assumptions about the initial conditions, such as the initial distance and the man's speed, which are not explicitly defined. The implications of the solutions derived from the equations are also not fully resolved, leaving open questions about the physical interpretation of the two times.

derekmohammed
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I was running for the bus when I thought this one up... (Maybe it is not so clever...)

Ok. A man is running at his top speed for the bus that is stopped at a stop light. The light turns green and the bus accelerates away from him at an acceleration of a metres per second per second. At what time T does the man catch the bus (assuming that he does catch the bus). When you solve the problem you have two Times. Are both valid? If so what do they represent...

Have fun.
 
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There's a condition which has to be met,in order for the problem to have sense...Let's denote by x_{0} the distance between the man and the bus,when the bus begins 2 accelerate.Then,assuming the acceleration to be +1ms^{-2},the condition reads:
v_{man}^{2}>=2x_{0}.For the equality case,the problem admits uniques solution,the man reaches the bus after exactly 2 x_{0}.
For the ">" case,the problem admits 2 solutions.The man reaches the bus in the interval (x_{0},2x_{0}) and overtakes and later,because of the acceleration,the bus overtakes the man.
For the "<" case,the problem has no solution,the man never reaches the bus...

Daniel.
 
call the initial distance between man and bus x
then the distance between them as a function of time is
d = x + 0.5at^2 - vt

the question is when is this distance equal to zero, so we must solve:
0 = x + 0.5at^2 - vt

which gives:

t = \frac{v+\sqrt{v^2 - 2ax}}{a} \vee t = \frac{v-\sqrt{v^2 - 2ax}}{a}

since the acceleration is 1 m/s2 this gives:
t = v+\sqrt{v^2 - 2x} \vee t = v-\sqrt{v^2 - 2x}

v2 must be equal or larger than 2x if it is equal there is just one solution: t = v
if v2 is larger than 2x there are two solutions: t1 = v - something and t2 = v + something

Both are valid, you will reach the bus after t1 seconds. If you keep running you will get ahead of the bus, then some time later at t2 the bus will catch up with you again.
 
1.You're 2 hours too late...:-p
2.Your solution does not obey Brain Teaser Forum standards,namely hiding your sollution,correct or not,it doesn't matter.

To the OP.This is not a Brain Teaser,this is a 9-th grade problem...

Daniel.
 
1. I did not find your explanation very clear
2. You cannot make the tex white, and since it is not a real "puzzle" I thought this would be ok...
 
1.I think it was more than clear.Any more clear than that,and the kindergarten kids would have gotten it.
2.I knew that.:-p :-p

Daniel.
 
:-p Hey I never said it was that clever! :-p
 
Nobody said anything about "clever"...Just to tease the brain...:wink:

Daniel.
 

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