Bukhovtsev-et-al-Problems-in-Elementary-Physics

MECHANICS
s
165
J 2
B
j t; hours Fig. 270
It can
be seen from Fig. 269 that
t
BF
tan ~·t3-tan(Xl·t l
V2= an a.=n= t. 2.5 km/h
(2) From the moment the ship meets the rafts to the moment It overtakes
them, the rafts will cover a distance equal to
s=v. (t 1+t.+t 8 )
On the other hand, this distance is equal to the difference between the distances
travelled by the ship upstream and downstream:
Hence,
s=t 3 (Vi+V2)-
t 1 (VI-VI)
and
13. The motion of the launches leaving their landing-stages at the same
time is shown by lines MEB and KEA, where E is their meeting point
(Fig. 210). Since the speeds of the launches relative to the water are the same,
MA and KB are straight lines.
Both launches will travel the same time if they meet at the middle between
the landing-stages. Point 0 where they meet Iles on the intersection of
line KB with a perpendicular erected from the middle of distance KM. The
motion of the launches is shown by lines KOD and COB. It can be seen from
Fig. 270 that AM AF is similar to I1COF, and, therefore, the sought time
MC=45 minutes..
14. The speed of the launches with respect to the water VI and the velocity
of the river current v 2 can be found from the equations s= t1 (VI +VI) and
s=t 2 (V l - V 2
) , where it and t 2 are the times of motion of the launches downstream
and upstream. It follows from the condition that 11= 1.5 hours and
t,=3 hours.