254

THEORY AND PRACTICE

TRACT 37*

88. And conversely, to find what velocity will remainafter passing over any space, as 4000 feet, the first velocitybeing v = 200. Here the hyp. log. of — is =

*72178, the natural number of which is 2*06 nearly, that is,v » v 200

2 06 = — : therefore v = —— = —— = 97*1, the velocity.v 7 2'0G 2 - ou 7 J

89. Again, for the time t\ since x — —— x —there-° 7 32 etc* v 7

fore 1 — x = x the correct fluent of which is

v 32ac a u a

t =

x (-

' n

1

X V —f’.—So, for example, if200 _ 2

30000 ~~

VV

V — V

300

; and

32cd' 4 ^ v v * 32oi a

V = 300, and v ~ 100; then

~~r or 5542 x -- = 36"*95 = t, the time of reducing the300 velocity to 100, or of passing over the space 6108 feet.

90. And, reversing, to find the velocity v, answering to

any given time t : Since t = —~ X (— --) = 5542 X

--), therefore v = • Here, if t be given = 30".

(v v 17 5542+ tv 7 » »

5542v

and v = 300; then t> =velocity sought.

5542

= 17X17, x 300 = 114, the14542 ’

91. Corol. The same form of theorem, x ~ X h. I.

—, as above is brought out for small velocities, will alsoserve for the higher ones, if we employ the medium resist-ance between the two proposed velocities, as was done inprob. 5 . Tims, as in the first example of this problem,where the two velocities are 1780 and 1500, the resistancedue to the velocity 1700, in the table of resistances, being76-78, say as 1700 1 : 1780 *: : 76*78 : 84*18, the resistancedue to the velocity 1780; then the mean between 84'18 and59-20, due to 1500 velocity, is 71*69, or rather take 72.Again, as V 67‘93 *. V72 : : 1600 : 1G47, the velocity due tothe medium resistance 72. Hence, as in prob. 5, as 1647* *.v* : : 72 : -0000265411* = suppose av z , the resistance due toany velocity v, between 1780 and 1500, for the l|lb ball.