Solving for Closest Approach: 2D Kinematics Help and Solutions

[Kudo Shinichi]

Urgent!HELP!A problem on 2D kinematics

Homework Statement

A flock of Canada geese directly fly above Kingston is flying due north at a speed of 25km/hr. A second flock directly above Gananoque, 30km to the east of Kingston, is flying in a northwesterly direction at the same speed. Assume there is no wind where the geese are flying.
a)What is the distance of closest approach between the two flock
b)How long does it take to reach this point?
c)What are the position vectors of the two flocks relative to Kingston at this instant of time?

Can anyone help me? thank you very much.

The Attempt at a Solution

The formula for distance is ((30-25t/sqrt(2)) - 0)^2 + (25t - 25t/sqrt(2)) ^2 = 625t^2(3/2 - sqrt(2) + 1/4) -25 t sqrt(2) + 900
Derivative: 625t (3.5 - sqrt(2)) -25 sqrt(2). =0
t = 1/25(3.5 - sqrt(2)).

 

[Doc Al]

Looks like the right approach.

The Attempt at a Solution

The formula for distance is ((30-25t/sqrt(2)) - 0)^2 + (25t - 25t/sqrt(2)) ^2

Looks good.

= 625t^2(3/2 - sqrt(2) + 1/4) -25 t sqrt(2) + 900

Redo this step. I get a different answer.