Engineering Mechanics

Chapter 19 : Relative Velocity „„„„„ 415

Solution. Given : Velocity of ship A = 12 km.p.h. (North) ; Velocity of ship B = 20 km.p.h.
(North-East) ; Distance between the ships at midnight = 40 km and vision distance = 10 km

Fig 19.17.
First of all, let us draw the actual velocity diagram of the two ships A and B as shown in Fig. 19.17
(a). Now draw the relative velocity diagram as shown in Fig. 19.17 (b) and as discussed below :

1. First of all, draw a line XL towards North representing the actual direction of motion of
   the ship A with a velocity of 12 km.p.h.


2. Now cut off XM equal to 12 km to some suitable scale towards South i.e. on the opposite
   direction of the actual motion of the ship A.


3. Now cut off XN equal to 20 km to the scale towards North-East, which represents the
   actual velocity of the ship B.


4. Complete the parallelogram XMRN with XM and XN as adjacent sides.


5. Join XR, which gives the magnitude and direction of the relative velocity. By measurement,
   we find that XR = 14.3 km.p.h.
   Now draw the exchange of signal diagram for the two ships with the help of their relative
   velocity diagram as shown in Fig. 19.18 and as discussed below :

Fig. 19.18. Exchange of signal diagram

1. First of all, draw the relative velocity diagram and the parallelogram XMRN as discussed
   above, and extend the line XR to T.


2. At midnight, let the ship A be at X. Therefore the ship B will be 40 km towards East. So
   mark XQ equal to 40 km to some scale, which represents the position of the ship B at
   midnight.


3. With Q as centre and radius equal to 10 km to the scale (i.e.vision distance) draw an arc
   meeting the relative velocity line at P 1 and P 2. By measurement, we find that XP 1 = 31.5
   km and XP 2 = 47.5 km.


4. Thus the two ships start signalling from
   31.5
   2.20
   14.3

= hours to

47.5

3.32

14.3

= hours after

midnight. Therefore the two ships will continue signalling for 3.32 – 2.20 = 1.12 hour

= 1 hour 7 min 12 sec. Ans.