One ship is sailing South at the rate of 5 knots, and another is sailing East at a rate of 10 knots. At 2 p.m the second ship was at the place occupied by the first ship one hour before. At what time was the distance between the ships not changing?
at 1 PM the 1st ship is 10 nautical miles due east of the 2nd
at 2 PM the 1st ship is 5 nautical miles due south of the 2nd
the distance (x) between them [at time (t)] is ... √[(10 - 10t)^2 + (5t)^2]
x = (125 t^2 - 200 t + 100)^(1/2)
find dx/dt and set it equal to zero
... solve for t
how did it became 5t+5?
at 2 pm say ship a is at the origin and sailing east at 10 kn
at 2 pm ship b is at (0, -5) and sailing south at 5 kn
say time t is time after 2 pm
xa = 10 t
yb = -5 - 5 t
if s is separation distance
s = sqrt(xa^2 + yb^2)
s^2 = xa^2 + yb^2
2 s ds/dt = 2 xa dxa/dt + 2 yb dyb/dt
when is ds/dt = 0?
xa dxa/dt = - yb dyb/dt
10 t *10 = -(-5-5t)(-5)
100 t = -25 -25 t
125 t = -25
t = -1/5 hour
12 minutes before 2
If we set t=0 at 2 pm, we have a right triangle at time t hours later, such that the distance z between the ships is
z^2 = (5(t+1))^2 + (10t)^2
= 125t^2 + 50t + 25
if we want z not to be changing, then we need
dz/dt = 0
2z dz/dt = 250t+50
dz/dt = 50(5t+1)/z
clearly dz/dt=0 when t = -1/5
That is, at 1:48 pm the distance was at a minimum.
To determine the time when the distance between the ships was not changing, we need to find the point where the two ships' paths intersected.
Let's first establish a coordinate system to help visualize the situation. We can assume the starting point of the first ship as the origin (0,0), with the positive x-axis pointing East and the positive y-axis pointing North.
Ship 1 is sailing South at a rate of 5 knots. Therefore, its position after t hours can be represented as (0, -5t).
Ship 2 is sailing East at a rate of 10 knots. Therefore, its position after t-1 hours (since it was at the place occupied by the first ship one hour before) can be represented as (10(t-1), 0).
To find the time when the ships' positions intersect, we need to equate their x and y coordinates:
0 = 10(t-1) ....(1) (Equating x coordinates)
-5t = 0 ....(2) (Equating y coordinates)
From equation (2), we can solve for t:
-5t = 0
t = 0
Plugging this value of t into equation (1):
0 = 10(0-1)
0 = 10(-1)
As we can see, there is no solution for this equation. This means that the two ships' paths do not intersect at any point, and therefore the distance between them does not remain the same at any time.
Hence, the distance between the ships is never not changing.