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

166
ANSWERS AND SOLUTIONS
Hence
and therefore
5 (t 1+12 ) 15 k h
VI = 2/
1
t 2
m/
5 (t -t )
V 2 = 2 1 = 5 km/h
2t 1 t 2
The point of the meeting is at a distance of 20 km from landing-stage M.
15. Let us assume that the river flows from C to T with a velocity of V o
Since the duration of motion of the boat and the launch is the same, we can
•
write the equation
_5=2(_5+_5)
VI + Vo V2+ Vo v 2 - Vo
where s is the distance between the landing-stages. Hence,
v~+4v2Vo+4V 2 V1 - v: = 0
Vo= -20 2 ± V 5v:-4vIV2= -20 ± 19.5
The solution Vo = - 39.5 krn/h should he discarded, since with such a current
velocity neither the boat nor the launch could go upstream.
For this reason. Vo=- 0.5 krn/h, i. e.• the river flows from T to C.
16. The speed of the boat v with respect to the bank is directed along AB
(Fig. 271). Obviously, v=vo+u. We know the direction of vector v and the
magnitude and direction of vector Yo. A glance at the drawing shows that
vector u will be minimum when u1-v.
Consequently.
b
Umin =Vo cos a. where cos a
Ya 2 +b 2
17. Let the speed u be directed at an angle a. to the bank (Fig. 272). Hence
t (u cos a-v)=BC=a. and tu sin a=AC=b
where t is the time the boat is in motion.
By excluding a from these equations. we get
t 2 (u 2_v2)-2vat-(a2+b2)=O
whence t= 15/21 hour. It is therefore impossible to cover the distance AB in
30 minutes.
c a B
c
6
A
Fig. 271
Fig. 272