Question

Ships A and B leave port together. For the next two hours, ship A travels at 40.0 mph in a direction 35.0° west of north while the ship B travels 80.0° east of north at 20.0 mph . What is the distance between the two ships two hours after they depart? What is the speed of ship A as seen by ship B?

Answer 1

Answer:

Explanation:

Given

Ship A velocity is 40 mph and is traveling 35 west of north

Therefore in 2 hours it will travel $40\times 2=80 miles$

thus its position vector after two hours is

$r_A=-80sin35\hat{i}+80cos35\hat{j}$

similarly B travels with 20 mph and in 2 hours

$=20\times 2=40 miles Its position vector[tex]r_B=40sin80\hat{i}+40cos80\hat{j}$

Thus distance between A and B  is

$r_{AB}=\left ( -40sin80-80sin35\right )\hat{i}+\left ( 80cos35-40cos80\right )\hat{j}$

$|r_{AB}|=\sqrt{\left ( -40sin80-80sin35\right )^2+\left ( 80cos35-40cos80\right )^2}$

$|r_{AB}|=103.45 miles$

Velocity of A

$v_A=-40sin35\hat{i}+40cos35\hat{j}$

Velocity of B

$v_B=20sin80\hat{i}+20cos80\hat{j}$

Velocity of A w.r.t B

$v_{AB}=v_A-v_B$

$v_{AB}=\left ( -20sin80-40sin35\right )\hat{i}+\left ( 40cos35-20cos80\right )\hat{j}$