MHB Distance Travelled: Equation for a,j,p - Solved

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The discussion centers on deriving the equation for the distance traveled by Jing and Pieter when they meet. It establishes that the distance Jing rides to catch up is represented by the formula d = aj/(j-p), where 'a' is the initial distance between them, 'j' is Jing's speed, and 'p' is Pieter's speed. The relationship between their travel times is based on the equation d-a = distance Pieter rides, leading to the conclusion that both riders cover the same distance when they meet. The analysis includes setting up equations based on their respective speeds and distances from Pieter's house. Ultimately, the thread resolves the problem using algebraic manipulation to find the distance Jing travels.
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The answer is ja/j-p kms but I have no idea how to get there
 

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let $d$ be the distance Jing rides to the point of catching up

$d-a$ = distance Pieter rides to the same point

$\dfrac{distance}{rate} = time$

time Pieter rides = time Jing rides

$\dfrac{d-a}{p} = \dfrac{d}{j}$

$dj - aj = dp$

$dj - dp = aj$

$d(j-p) = aj$

$d = \dfrac{aj}{j-p}$
 
Another way: Pieter rides west at p km/hr so after 't' hours he will be pt km from his house.

Jing Jing rides at rides west at j km/hr so after 't' hours he will be jt km from his house. But Jing Jing started "a" km to the west of Pieter's house so after 't' hours Jing Jing will be jt+ a km from Pieter's house.

They meet when they are at the same place so the same distance from Pieter's house: jt= pt+ a. Subtract pt from both sides: jt- pt= a. Factor t out on the left: (j- p)t= a. Divide both sides by j- p: t= a/(j- p).

Jing Jing was riding at j km/hr so he will have ridden aj/(j- p) km.
 
...aNUTter look-see:

[P]@p...pt...>t

[J]@j....pt+a....>t
 
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