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

  • Thread starter derekmohammed
  • Start date
  • Tags
    Bus
In summary, the conversation involves a man trying to catch a bus that is stopped at a stop light. The problem is solved by calculating the distance between the man and bus, and solving for when it will be equal to zero. This results in two valid solutions, with the possibility of the man catching up to the bus and then being overtaken by it again. However, there is a condition that must be met for the problem to make sense.
  • #1
derekmohammed
105
0
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.
 
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • #3
call the initial distance between man and bus x
then the distance between them as a function of time is
[tex]d = x + 0.5at^2 - vt [/tex]

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

which gives:

[tex] t = \frac{v+\sqrt{v^2 - 2ax}}{a} \vee t = \frac{v-\sqrt{v^2 - 2ax}}{a}[/tex]

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

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.
 
  • #4
1.You're 2 hours too late...:tongue:
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.
 
  • #5
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...
 
  • #6
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.:tongue: :tongue2:

Daniel.
 
  • #7
:tongue: Hey I never said it was that clever! :tongue:
 
  • #8
Nobody said anything about "clever"...Just to tease the brain...:wink:

Daniel.
 

What is "Solve for T: Catch the Bus!"?

"Solve for T: Catch the Bus!" is a mathematical problem that involves finding the value of "T" in an equation that represents the time it takes to catch a bus. It is often used as an example in algebra or physics classes to demonstrate the application of equations in real-life scenarios.

What is the equation used to solve for T in "Solve for T: Catch the Bus!"?

The equation used in "Solve for T: Catch the Bus!" is T = D/V, where T represents the time, D represents the distance, and V represents the speed of the bus.

What information is needed to solve for T in "Solve for T: Catch the Bus!"?

To solve for T in "Solve for T: Catch the Bus!", you will need to know the distance between the starting point and the bus stop, as well as the speed of the bus. These values can be given in any unit of measurement, as long as they are consistent.

What are some real-life applications of "Solve for T: Catch the Bus!"?

The concept of "Solve for T: Catch the Bus!" can be applied in various real-life scenarios, such as calculating the time it takes to reach a destination by car, or determining the time it takes to catch a train or flight. It can also be used in physics to calculate the time it takes for an object to reach a certain distance with a given velocity.

What are some tips for solving "Solve for T: Catch the Bus!"?

When solving "Solve for T: Catch the Bus!", it is important to make sure that the units of distance and speed are consistent. If they are not, you will need to convert them to the same unit before solving the equation. It is also helpful to draw a diagram or visualize the scenario to better understand the problem.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Special and General Relativity
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
930
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
10
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
4K
Back
Top