If we call A's position at noon (0,0) then we have the positions of A and B after t hours
A: (12t,0)
B: (0,60-9t)
The distance is thus
d^2 = (60-9t)^2 + (12t)^2 = 225t^2-1080t+3600
At t=2, d=√(42^2+24^2)=√2340=48.37
So, at any time t,
2d dd/dt = 450t-1080
Now just plug in your values
At noon of a certain day,ship A is 60 km due north of ship B. If A sails east at 12 km/hr and B sails north 9 km/hr,determine how rapidly the distance between them is changing 2 hrs later. Is it increasing or decreasing?
2 answers
2d dd/dt = 450t-1080
dd/dt = 900-1080/(2*48.37)=-1.86
this is what I get when I substitute the values.Is it correct?
dd/dt = 900-1080/(2*48.37)=-1.86
this is what I get when I substitute the values.Is it correct?