......`\
%
/ (^) B
1 i
s i
p i I
% I 1
1
/ 1
A
r
—,.
-/ — —
. 13
1
1 e
8 '''',... ...7" ......_ _...."
CI
,,___....
C
Fig. 135
\
t
p
;
I
I
V toe <Owind
I
I
I
Vloc> vwind
,
Fig. 137
Solutions 125
of velocities vi oc and vw ind, the shape of the lower
part of the path will be different. It is a loop
(when vim > vwind, Fig. 136), or a parabola (when
Pi0C < vwind, Fig. 137), or a "beak' (when vion
=
vwind, Fig. 138). The latter case corresponds to the
Fig. 139
condition of our problem. Thus, the velocity of the
wind is vwind = vie = 10 m/s, and in order to find
its direction, we draw the tangent CC' from the
point of the "beak" to the path of the locomotive
(viz, the circle) relative to the ground.
1.23*. Let the merry-go-round turn through a cer-
tain angle ,:p (Fig. 139). We construct a point 0