# Thread: direction of a boat

1. ## direction of a boat

a boat A is heading south at 30 km/hr. its spotted a boat B which is 18 km due east of boat A. boat B has a max speed of 25 km/hr and wishes to get a close as possible to boat A. what direction should boat B head in.

so i initially let boat A be the origin heading south. boat B starts at (18,0) and heads south of west. so i let 18-25sina=0 solving this gives 46 degrees. the answer is 56.4 south of west. i can explain what i did in more detail if anyone needs it.

2. ## Re: direction of a boat

18 is a distance (km)
25 is a velocity (km/h) and 25sina is a component of that velocity.

Therefore you can't have 18-25sina because the units of the terms are different. ie you can't subtract km/h from km.

3. ## Re: direction of a boat

Oh ye. Any ideas?

4. ## Re: direction of a boat

I arrived at $\theta = \arcsin\left(\dfrac{5}{6}\right)$ degrees S of W for boat A's course, but I had to resort to minimizing with partial derivatives to arrive at that conclusion. Must be an easier way, but I can't see it.

5. ## Re: direction of a boat

I am squaring the distance in the y direction between the boats and I am squaring the distance in the x direction between the boats and letting it equal to zero because the the boat B is aiming to intercept boat A.

i get (25cos(a)t-18)^2 + (30t-25sin(a)t)^2 =0 i trying to work this out. is it possible? im trying to get rid of t.

also i was thinking an easier way to solve it would be to say the distance travelled by the boats in the y direction is equal so 25sinat=30t so 25sina=30 sina=30/25 sadly this doesnt work. why doesnt it work do you know? however if i swap this around it does work sina=25/30 gives the right answer. is this coincidence or is there something wrong with the question

6. ## Re: direction of a boat

boat B never catches boat A ... the minimum distance will be > 0.