Engineering Mechanics

Chapter 19 : Relative Velocity „„„„„ 409

(b) Bearing of the submarine

Fig. 19.9.

1. Again draw East, West, North and South lines meeting at X.


2. Through X. draw a line XK at 35° with the East, which represents the actual direction of
   the enemy ship. Now cut off XN equal to 12.7 knots to some suitable scale.


3. Now cut off NQ equal to 15 knots to the scale (equal to the relative velocity of the submarine
   with respect to the warship). Join NQ.


4. Now cut off XR equal to 150 nautical miles on the line representing the South direction
   (because the warship is stationed 150 nautical miles south of the submarine) to some
   other scale. Through R, draw RS parallel to the line QN meeting the line XK at S as shown
   in Fig. 19.9.


5. By measurement, we find that bearing of the submarine ∠ XRS = 43° with North. Ans.

Time taken by the warship to intercept the enemy ship.

By measurement, we also find that distance RS = 123 km.

Therefore time taken by the warship to intercept the enemy ship

Distance 123

hours

Velocity of warship 18

RS

==

= 6.83 hours = 6 hours 50 min Ans.
Example 19.7. A rifleman on a train, moving with a speed of 52 km per hour, fires an object
running away from the train at right angle with a speed of 39 kilometres per hour. The line connecting
the man and the object makes an angle of 30° to the train at the instant of shooting.

At what angle to the train should he aim in order to hit the object, if the muzzle velocity is 200
metres per second?

Solution. Given : Speed of train = 52 km.p.h. ; Speed of object = 39 km.p.h. ; Angle, which
the line connecting the man and object makes with the train = 30° and muzzle velocity = 200 m/s