Two people start from the same point. One walks east at 1 mi/h and the other walks northeast at 2 mi/h. How fast is the distance between the people changing after 15 minutes?

Answer 1

At a time of t hours
let the distance covered by the eastbound person be 1t km
let the distance covered by the northeastbound person be 2t km
let the distance between them be d km
by the Cosine Law
d^2 = t^2 + 4t2 - 2(t)2t)cos45°
= 5t^2 - 4t^2(√2/2)
= 5t^2 - 2√2 t^2
when t = 15 min or .25 hours
d^2 = 5(.25)^2 - 2√2(.25)^2 = .489277
d = .6995

2d dd/dt = 10t -4√2t
dd/dt = (5t - 2√2t)/d
= (5(.25) - 2√2(.25))/.6995 = .776

at that moment they are separating at .776 km/h

check my arithmetic

Answer 2

Incorrect

Answer 3

If you had checked my arithmetic like I suggested you would have found my error to be in
d^2 = 5(.25)^2 - 2√2(.25)^2 = .489277
It should have been
d^2 = 5(.25)^2 - 2√2(.25)^2 = .135723
then
d = .3684

I will let you make the rest of the corrections.

BTW, I also used km/h instead of mi/h, but that has no effect on the calculation