Question 339836
you basically have a right triangle formed by the movement of the cars, one from coming from the east and one coming from the south
--
let d=ditance between the cars.
let x=distance from car from the east to the meeting point
let y=distance from car from the south to the meeting point
distance=speed*time; d=s*t
let {{{D[e]}}}=Distance by car from east
let {{{D[s]}}}=Distance by car from south
let {{{d[e]}}}="shrinking" distance by car from east
let {{{d[s]}}}="shrinking" distance by car from south
let {{{D[c]}}}=Distance between cars
let {{{d[c]}}}="shrinking" distance between cars
--
at t=0, one car is 4 miles east and the other car is 3 miles south, so the distance between them, by the pythagorean theorem:
  {{{D[c]=sqrt(x^2+y^2)=sqrt(3^2+4^2)=5}}}
--
as time goes forward, the distances from the east and the south shrink toward their meeting point.
car from the east travels a distance:  {{{D[e]=50*t}}} in general
relative to the meeting point 4 miles away the distance shrinks
as {{{d[e]=4-50*t}}}
--
car from the south travels a distance:  {{{D[s]=40*t}}} in general
relative to the meeting point 3 miles away the distance shrinks
as {{{d[s]=3-40*t}}}
--
{{{d[c]=sqrt(d^2[e]+d^2[s])=sqrt((4-50*t)^2+(3-40*t)^2)}}}