Asked by Michael
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?
Answers
Answered by
Reiny
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
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
Answered by
Michael
Incorrect
Answered by
Reiny
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 = <b>.135723</b>
then
d = <b>.3684</b>
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
d^2 = 5(.25)^2 - 2√2(.25)^2 = .489277
It should have been
d^2 = 5(.25)^2 - 2√2(.25)^2 = <b>.135723</b>
then
d = <b>.3684</b>
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