# Thread: relative velocity and interception

1. ## relative velocity and interception

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 .

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

course of A - 283.89 degrees .

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

2. 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 .

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

course of A - 283.89 degrees .

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

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

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

In the interception phase, the velocity of A relative to B is (according to my working, and I think to yours) $-48.54\textbf i$. Therefore the time A takes to intercept B is:
$\frac{10}{7\times 48.54}$ hours $= 1.77$ minutes
But check my working, won't you?

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

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

In the interception phase, the velocity of A relative to B is (according to my working, and I think to yours) $-48.54\textbf i$. Therefore the time A takes to intercept B is:
$\frac{10}{7\times 48.54}$ hours $= 1.77$ minutes
But check my working, won't you?

hello grandad , i got 2.58 minutes . Err , could you check my working for my mistake ?

The ship move to A' when its exactly east of B , and since the course changes , Va=xi+yj

velocity of A relative to B= Va-Vb = xi+yj-(15i+12j)

b-12=0 , b=12 (vector j is 0)

|Va|=50

$\sqrt{a^2+b^2}=50$

a=48.54

so Va=48.54i+12j

so direction would be N 76.1 E

the shortest distance is 1.26 km , sin 5.56=d'/13 ,d'=1.43

aVb=33.54

s=vt , 1.43=33.54t , t=0.043 hours = 2.58 mins .

4. Hello hooke
Originally Posted by hooke
hello grandad , i got 2.58 minutes . Err , could you check my working for my mistake ?

The ship move to A' when its exactly east of B , and since the course changes , Va=xi+yj

velocity of A relative to B= Va-Vb = xi+yj-(15i+12j)

b-12=0 , b=12 (vector j is 0)

|Va|=50

$\sqrt{a^2+b^2}=50$

a=48.54

so Va=48.54i+12j

so direction would be N 76.1 E

the shortest distance is 1.26 km , sin 5.56=d'/13 ,d'=1.43

aVb=33.54

s=vt , 1.43=33.54t , t=0.043 hours = 2.58 mins .
The shortest distance, $1.26$ km, has nothing to do with it. Ship A changes course when it is due East of B, not when it's at the shortest distance from B.

When A is due East of B, then, as I showed before, it is $\tfrac{10}{7}= 1.429$ km away, and its velocity relative to B is then $-48.54\textbf i$ (which you have confirmed in your working). So this distance $1.429$ km is being 'closed down' at a rate of $48.54$ km per hour. So the time A takes to intercept B is
$\frac{1.429}{48.54}=0.0294$ hours
$=1.77$ minutes