Relative velocities, and positions

[d=vt+1/2at^2]

Homework Statement

-Ship A is located 4 km north and 2.5 km east of ship B
-ship A has a velocity of 22km/h south
-ship B has a velocity of 40km/h in a direction 37 degrees north of east

a) Find the velocity of A relative to B in unit vector notation, with i towards the east
b) Write an expression in terms of i and j for the position of A relative to B as a function of t, where t = 0 when the ships are in the positions described above
c)at what time is the separation between the ships least?
d) What is that least separation?

Homework Equations

The Attempt at a Solution

I believe the answer to part a, is V(A) = -22j

I have absolutely no idea how to do b, c, or d

Please do not solve the question for me, but if someone could point me in the right direction, it would be GREATLy appreciated!

 

[latrocinia]

a) you have to divide to vectors in different components, so for example the vector that describes the motion of A = (0i, -22j). You should do the same for B.

b) if you have the relative velocity we can assume that one of the ships doesn't move (which is the definition of 'relative motion'). so you begin with the initial distance between the ships and you this is becoming less by the second with the relative velocity.

 

[d=vt+1/2at^2]

Ok, I got a and b, and now I am stuck on C. How do I find the time when the separation of the ships is the least?

 

[latrocinia]

i don't know which formula you have fabricated at b. but if b is correct you can do two things to solve c.

1) you can plot it with on the y-axis the distance between the ships (delta x) and on the x-axis the time. this way you can see at what time the minimum occurs.

2) you can differentiate the function you found at b ( dx/dt). for the minimum let this equal to 0. since then you have a maximum or minimum found in your function. but i don't know if you are advanced enough to know this technique.

 

[d=vt+1/2at^2]

My formula for B is correct, i tested it out on graph paper.

I got: -40cos37it - 46.1jt + 2.5i + 4j

To summarize, after one hour (since the velocities are in km/h), relative to ship B, ship A is (-40cos37+2.5) units West of ship B, and (-46.1+4) units south of ship B.