Question

One car leaves a station going west at 90 kph. Another approaches the same station from the south at 70 kph. At what rate is the distance between the cars changing when the cars are each 20 km from the station?

Answer 1

The triangle is isoceles, and what we need is the projection of the velocities on the line that aonnect both cars.
Since the angles are 45° the rate you want is: 90*cos(45°)+70*cos(45°)

Answer 2

Ha, that seems soo simple using trig functions compared to using Pythagorean's theorem

Answer 3

from what i have been working on trying to us the pt i have a=20 da/dt=90 b=20 db/dt=-70 c=√800 does that seem right @Chlorophyll ?

Answer 4

but when i solve for dc/dt i come out with 14.142 which seems a little off to me

Answer 5

well actually that wouldnt be too illogical with the distance changing at 14 km/hr

Answer 6

z² = x² + y²
=> z = √ 90 ² + (-70) ² = 114.02

-> zz' = x x' + y y'
= 20 ( 20 ) = 400 => z' = 400 / 114.02 ≈ 3.51 km

Answer 7

According to your sample, approaching the intersection is negative sign!
( That's kind of new to me, as I thought normally applying the quadrant for the sign sine, cosine Hmm)

Answer 8

z' = 3.51 kph ( Look like the second one falls back behind )

Answer 9

@sjerman1 's answer is correct, it is 14.142, not 3.51.
Using Pythagorean theorem is easy, you just need to identify your triangle correctly.
You want how fast the second car is going in the direction of the first, so:
Doing the same for the other car, you get:
v1^2=2vf1^2
vf1=v1/sqrt(2)
so the rate that the distance between them is changing is vf1-vf2=(90-70)/sqrt(2)=14.142