Relative motion in two dimensions question

[f25274]

Ship A is located at 4 km north and 2.5 km east of ship B. Ship A has a velocity of 22 km/h toward the south, and ship B has a velocity of 40 km/h in a direction 37 degrees north of east. (a) What is the velocity of A relative to B in unit-vector notation with i toward 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

i,j are unit vectors.
v_PA=V_PB+V_BA

The Attempt at a Solution

Angles are in degrees
(a)
dr/dt_A= -22km/h j
dr/dt_B= 40cos(37)i+40sin(37)j
dr/dt_BA= -40cos(37)i-22km/h j-40sin(37)j
(b)
$\int$dr_BA=$\int$-40cos(37)i-(22km/h+40sin(37))j dt
r_BA=(2.5-40cos(37)t)i+(4-t(22+40sin37))j
(c)
Now I don't know how to solve it lol
(d)

I'm really lost on how you can solve the minimum distance or the time when it reaches the minimum distance.
Please give me at least a hint

 

[Doc Al]

(a)
dr/dt_A= -22km/h j
dr/dt_B= 40cos(37)i+40sin(37)j
dr/dt_BA= -40cos(37)i-22km/h j-40sin(37)j

OK, but why not just symbolize the velocities as V_A and V_B?

(b)
$\int$dr_BA=$\int$-40cos(37)i-(22km/h+40sin(37))j dt
r_BA=(2.5-40cos(37)t)i+(4-t(22+40sin37))j

The velocities are constant, so get rid of the integral sign. (Do the integration, if you must.)

(c)
Now I don't know how to solve it lol

First get a simpler expression for part b. Then how would you write the separation as a function of time? Then you can finally use some calculus.

 

[f25274]

I don't think I understand what you mean by separating as a function of time
r_BA=(2.5-40cos(37)t)i+(4-t(22+40sin37))j
r_BA=(2.5-31.945t)i+(4-46.073t)j

:( Is this what you mean by simple?

 

[Doc Al]

I don't think I understand what you mean by separating as a function of time
r_BA=(2.5-40cos(37)t)i+(4-t(22+40sin37))j
r_BA=(2.5-31.945t)i+(4-46.073t)j

:( Is this what you mean by simple?

I'm sorry... you did do the integration. My bad!

The separation is the distance. Now that you have the relative position, how can you express the distance as a function of time? Hint: What's the distance squared?

 

[Lobezno]

As always when confused, start by drawing a graph. Let's call the position of ship A at time t=0 (0,0,0). That will make the vectors fixed.