MATHEMATICAL REPRESENTATION OF VECTOR 489 



Now ^= VA a +B a (Ycos . A -sin B 

 di \/A 2 ( B 2 



- A/A 2 + B 2 {(cos cos a - sin sin a) 

 + i (sin cos a + cos sin a) } 



= V A 2 + B 2 {cos (0 + a) -i i sin (0 + a) } where tan a = -r 



and this gives the acceleration in the recognised form for a vector. 



Hence if a velocity is a vector of magnitude p and direction 0, 



then the corresponding acceleration is also a vector of magnitude 



^dp\ z TdOX 2 , dQ 



~T i + I P ~37 ) an " direction (0 + a) where tan a = p -T: / 

 dt / \ dt / di / dp 



The angle a is evidently the angle between the direction of the 

 velocity and the direction of the acceleration corresponding to 

 that velocity. 



Taking the case of a body describing a circular path, of radius 

 r feet, with uniform velocity v feet per second. This velocity 

 can be expressed as a vector of magnitude p and direction 



j 



and p = v a constant and = . 



r 



Then -37 = and -3- = - 



dt dt r 



also p -T- = 



1 at r 



Now the velocity = p (cos + i sin 0) 



/ /^7 \ 2 / Jfl\ 9 



and the acceleration - A/(T J + ( P-JI ) ( cos (0 + a) + f sin (0 + a) } 

 ' \ai/ \ at/ 



Now 



and tan a = o -rr 



-rr / 

 a/ / dp 



" 



= oc since - = 



Hence a = 90. 



r 2 

 Thus the acceleration - -{cos (0 + 90) + i sin (0 +90) } 



