Hello hooke Originally Posted by

**hooke** Ship A sails with velocity (30i+40j) km/h and ship B sails with velocity (15i+12j) km/h . At one instant , the position vector of ship B relative to ship A is (5i+12j) km . Calculate the magnitude and direction of the velocity of ship A relative to ship B . Calculate also the shortest distance between the 2 ships .

My answers :

I found the magnitude of V of A relative to B - 31.76 km/h with a direction of N28.18 E

and the shortest distance - 1.26 km

Am i correct ?

This is the continuation of the question . At the instant when ship A is east of ship B , ship A changes its course but maintain its speed in order to intercept ship B . Find the course that ship A must steer and the time taken to intercept ship B

my answers :

course of A - 283.89 degrees .

time - i am not really sure . I don know the distance between A and B here .

I agree with all your answers. Good work!

In the first stage, the velocity of A relative to B is $\displaystyle 15\textbf i + 28\textbf j$, and the initial displacement of A from B is $\displaystyle -5\textbf i -12\textbf j$.

When A is due East of B, its displacement from B is $\displaystyle r\textbf i + 0\textbf j$, for some $\displaystyle r$. Considering the change in the $\displaystyle \textbf j$ component, the time taken for this is $\displaystyle \frac{12}{28} = \frac{3}{7}$ hours. In this time, the increase in the $\displaystyle \textbf i$ component of A's position relative to B is $\displaystyle \frac{3}{7}\times15=\frac{45}{7}$. So its position relative to B is now $\displaystyle \left(-5+\frac{45}{7}\right)\textbf i= \frac{10}{7}\textbf i$.

In the interception phase, the velocity of A relative to B is (according to my working, and I think to yours) $\displaystyle -48.54\textbf i$. Therefore the time A takes to intercept B is:

$\displaystyle \frac{10}{7\times 48.54}$ hours $\displaystyle = 1.77$ minutes

But check my working, won't you?

Grandad