At noon, ship A is 550 km west of ship B. Ship A is sailing east at 50 km/h and ship B is sailing north at 45 km/h. How fast (in km/h) is the distance between the two ships changing at 5:00 P.M.?
2007-02-25 20:00:23 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Geography
At noon: A=(0,0), B=(550,0).
As time (t) goes by: A=(45*t,0), B=(550,50*t)
(The distance of two points A(x1,y1),B(x2,y2) is:
dist(A,B)=sqrt((x1-x2)^2+(y1-y2)^2))
The distance of those 2 as time goes by:
d(t)=sqrt((45*t-550)^2+2500*t^2)=
sqrt(2025*t^2+302500-49500*t+2500*t^2)=
sqrt(4525*t^2-49500*t+302500).
The rate at which this distance changes is:
d'(t)=(sqrt(4525*t^2-49500*t+302500))'=
(4525*t^2-49500*t+302500)'/sqrt(4525*t^2-49500*t+302500)=
(9050*t-49500)/sqrt(4525*t^2-49500*t+302500)
After 5 hours the rate at which the distance changes will be:
d'(5)=(9050*5-49500)/sqrt(4525*5^2-49500*5+302500)=
(-4250)/sqrt(168125)=-4250/(25*sqrt(269))=170/sqrt(269) which is about -10.365 km/h which means that the distance is decreasing at that time.
2007-02-25 20:39:51 · answer #1 · answered by costasgr43 2 · 0⤊ 0⤋
calculate their relative positions at 5 pm then the rate of change is solved by calculating the rate of change of the sum of their velocities. It is a basic right angle vector problem. the distance between them being the hypotenuese of the triangle
2007-02-25 20:13:24 · answer #2 · answered by U-98 6 · 0⤊ 0⤋
95 km per hour
2007-02-25 20:11:27 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Hay Dude wrong class, this is Geography 101
Math is down the hall to the left!
2007-02-28 02:31:37 · answer #4 · answered by Anonymous · 0⤊ 0⤋
This is a problem in Navigation. I will not solve it for you but if you take a nautical map and use compass and ruler you the answer will be obvious.
2007-02-25 20:07:44 · answer #5 · answered by The Stainless Steel Rat 5 · 0⤊ 0⤋
Medford, Oregon.
2007-02-28 11:26:32 · answer #6 · answered by Anonymous · 0⤊ 1⤋