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

[Yazan975]

View attachment 8404

The answer is ja/j-p kms but I have no idea how to get there

 

[skeeter]

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

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

$dj - aj = dp$

$dj - dp = aj$

$d(j-p) = aj$

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

 

[HOI]

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.

 

[Wilmer]

...aNUTter look-see:

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

[J]@j....pt+a....>t