231 



ever, gives rather too small a result, as the average velocity must be under 

 87 feet a second, and I will try a more correct way of arriving at the result. 



Let w be the weight of the bird in lbs., V its velocity at starting, and v 

 its velocity after having sailed over s feet in t seconds ; and let v' be its 

 velocity after having sailed over s' feet in t' seconds. 



If now we suppose the time between t' and t to be very short we may 

 assume the resistance of the air to be constant throughout the small space s'— s, 

 and to be equal to x Av 2 . 



Therefore W „ W , 9 , A „ , , > 



— \ z = — v L + x Ax 1 (s — s). 

 2 2 2g v ; 



Or" • ^1 



Y. (v'2 _ v 2 )= -X Av 2 (s' 



2 o - • 



-B) 



W dv 2 , „ 



2g ds 





• dv2 - - 2 Fv 2 

 ds 





where F represents & ~ • 



.-. v 2 =:Ce- 2Fs 

 Now when s=o, v=V .-. C=V 2 



And we obtain v 2 =V 2 e— 2Fs 

 v=Ye- Fs 



e?s=~. . . (1) 



But when the time is very short we may suppose that the velocity of the 

 bird would remain the same throughout it, and therefore 



s'— s=v (if— t) 



Or ^=v=Ve- Fs 

 dt 



«•. FVt=e Fs + c 



Now when t=o, s==o . \ c= —1 



Ande Fs =FVt+l , . (2) 



Equating this result with that obtained in equation (1) we get,, 



V 

 L=FVt + l 



V 





F=(I- 



V V 



1 )~ 



/ vt 



_V-v 

 Vvt 



Substituting *5 — . 



for F v 



gAx 



V-v 



W " 



' Vvt 



w 



''• X= Vv 



(V-v) 

 tgA 



