At 12 noon the position vectors r and the velocity vectors v of two ships A and B are
rA = (2i+j)km vA = (3i+j)kmh^-1
rB = (-i-4j)km vB = (11i+3j)kmh^-1
(a) Show that at time t h after noon the position vector of B relative to A is given by:
[(8t-3)i+(2t-5)j]km
(b) Show that the distance d km between the vessels, at this time, is given by
d^2=68[t^2 - t+1/2]
(c) Hence show that the ships are nearest together at 12.30 p.m and find the distance between them at this time.
Hello, outlaw!
In hours, the position vectors are:At 12 noon the position vectors and the velocity vectors of two ships A and B are:
. . .
(a) Show that hours after noon, the position vector of relative to is given by:
. . .
. .
Then: .
Therefore: .
(b) Show that the distance between the vessels at time
. . . is given by: .
Therefore: .
We have .(c) Hence show that the ships are nearest together at 12:30 pm
and find the distance between them at this time.
Note that is at a minimum when is a minimum.
The graph for is an up-opening parabola.
. . Its minimum occurs at its vertex.
The vertex is at: .
We have: .
. . Hence, the vertex is at: .
Since the distance is a minimum when ,
. . the ships are at their nearest at
When
. . Therefore, their minimum distance is .