Solve for Time When Two Boys Meet While Moving - Travel Problem

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The discussion focuses on a travel problem involving two boys, Jack and Tim, moving towards each other or in the same direction. The key equations derived for their meeting time are t = x/(a+b) when moving towards each other and t = -x/(a-b) when moving in the same direction. Participants emphasize the importance of understanding relative velocity and the correct application of kinematic equations. There is a request for clarification on methods and basic concepts, indicating some participants are struggling with the material. The conversation highlights the need for foundational knowledge in physics to solve such problems effectively.
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Homework Statement


Jack and Tim were standing in the main street. The distance between the boys was x metres. At the same moment the boys started to move. Jack was running at a m/s and Tim walked at b m/s.

Homework Equations


Find the time when the boys met if
a) they moved towards each other,
b) they moved in the same direction, walker in front.

The Attempt at a Solution


a) t=d/s
t=x/(a+b)
They will meet after x/(a+b) seconds.
 
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like in forst case you superimposed the velocity of a on b to find the vel. of a wrt to b

do the same in case (b) ... find the vel. of a wrt to b and do the math
 
use the simple concept of relative velocity. that should do
 
Where is the HELP ?!
I haven't understand anything what it said in these two replies...what is this 'wrt'
You haven't even said if I'm right for the first question...
 
chawki said:
I haven't understand anything what it said in these two replies...what is this 'wrt'
wrt = with respect to
You haven't even said if I'm right for the first question...
Your answer for the first question was correct. (Not sure how you got it though.)

Using relative velocity makes it easier, but not if you've not covered that concept.
 
Doc Al said:
wrt = with respect to

Your answer for the first question was correct. (Not sure how you got it though.)

Using relative velocity makes it easier, but not if you've not covered that concept.

Thank you for your help Doc Al...let's concentrate on the first question at the moment...Can you please tell me another method
 
chawki said:
Can you please tell me another method
Let's start by you explaining your method, since you got the right answer.
 
Doc Al said:
Let's start by you explaining your method, since you got the right answer.

I can't explain it...i copied it from another exercise that looked like this one.
Also i haven't practised for a long time..i think all i need is a law or a rule from which i can start...the other things will come..i'm sure
 
It's rather difficult to give advice since I have no idea what you know or don't know. Nonetheless...

One way to go is to write expressions for the position of each as a function of time. (Measure position from the same point, of course.) Then you can solve for when and where they meet. Be sure to give the speeds the correct sign, depending on their direction of travel.
 
  • #10
It's NOT difficult.
Just write the starting formula and i will remember how to do it...
I haven't done these exercises for about 10 years...
 
  • #11
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  • #12
Ok thanks :) i will try...
 
  • #13
xj = at+x0
xt = bt+x0

and then ?
 
  • #14
The distance between them is the difference of those two, xj- xt.
 
  • #15
chawki said:
xj = at+x0
xt = bt+x0

and then ?
Some comments:
(1) Careful with signs. When they move towards each other, their velocities have opposite signs.
(2) The starting point x0 is different for each. Measure the position of each from the same point: The initial position of Jack, for example.

Revise the equations accordingly. Then you can solve for the time when their positions are equal.
 
  • #16
i think i can do part B by revising another problem which i posted few days ago...

b) we write the equations with respect to Jack

xj = at
xt = bt-x

xj=xt
at=bt-x
t*(a-b)=-x
t = -x/(a-b) ...the time they will meet
 
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