Higher Engineering Mathematics

232 VECTOR GEOMETRY

opposite in direction tope. From the geometry

of this vector triangle:

|qp|=

√

(45^2 + 552 )= 71 .06 m/s

and argqp=tan−^1

(

55

45

)

= 50. 71 ◦

but must lie in the third quadrant, i.e., the

required angle is 180◦+ 50. 71 ◦= 230. 71 ◦.

Thus the velocity of carQrelative to carPis

71.06 m/s at 230.71◦.

Now try the following exercise.

Exercise 95 Further problems on relative

velocity

1. A car is moving along a straight horizontal
   road at 79.2 km/h and rain is falling vertically
   downwards at 26.4 km/h. Find the velocity of
   the rain relative to the driver of the car.
   [83.5 km/h at 71.6◦to the vertical]


2. Calculate the time needed to swim across
   a river 142 m wide when the swimmer can
   swim at 2 km/h in still water and the river is
   flowing at 1 km/h. At what angle to the bank
   should the swimmer swim?
   [4 min 55 s, 60◦]


3. A ship is heading in a direction N 60◦Eata
   speed which in still water would be 20 km/h.
   It is carried off course by a current of 8 km/h
   in a direction of E 50◦S. Calculate the ship’s
   actual speed and direction.
   [22.79 km/h, E 9.78◦N]

21.6 Combination of two periodic

functions

There are a number of instances in engineering and
science where waveforms combine and where it is
required to determine the single phasor (called the
resultant) which could replace two or more sepa-
rate phasors. (A phasor is a rotating vector). Uses
are found in electrical alternating current theory,
in mechanical vibrations, in the addition of forces
and with sound waves. There are several methods of
determining the resultant and two such methods are
shown below.

(i) Plotting the periodic functions graphically

This may be achieved by sketching the sepa-

rate functions on the same axes and then adding

(or subtracting) ordinates at regular intervals.

(see Problems 9 to 11).

(ii) Resolution of phasors by drawing or

calculation

The resultant of two periodic functions may be

found from their relative positions when the

time is zero. For example, ify 1 =4 sinωtand

y 2 =3 sin (ωt−π/3) then each may be repre-

sented as phasors as shown in Fig. 21.17,y 1

being 4 units long and drawn horizontally and

y 2 being 3 units long, laggingy 1 byπ/3 radians

or 60◦. To determine the resultant ofy 1 +y 2 ,y 1

is drawn horizontally as shown in Fig. 21.18 and

y 2 is joined to the end ofy 1 at 60◦to the hori-

zontal. The resultant is given byyR. This is the

same as the diagonal of a parallelogram which is

shown completed in Fig. 21.19. ResultantyR,in

Figs. 21.18 and 21.19, is determined either by:

(a) scaled drawing and measurement, or

(b) by use of the cosine rule (and then sine rule

to calculate angleφ), or

Figure 21.17

Figure 21.18

Figure 21.19