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

MECHANICS 169
20. Let us use the following notation; u 12=speed of the second vehicle
with respect to the first one; U 21 =speed of the first vehicle with respect to
the second one.
Obviously, U 12=U21 and u~2=v~+v~+2vIV2 cos a (Fig. 275). The time
sought is t = 3- .
U 12
21. During the time ~t the straight line AB will travel a distance ol~t
and the straight line CD a distance u2~t. The point of intersection of the
lines will travel to position 0' (Fig. 276). The distance 00' can be found
from triangle OFO' or OEO', where OF= v~ At =£0' and OE= v~ At =FO', i. e.,
sin a sin ~
whence
00' = YOF2+OE2+20F· OE cos ~=V &t
1-2. Kinematics of Non-Uniform and Uniformly
Variable Rectilinear Motion
22. The mean speed over the entire dis lance Om = t1+ :2 + t3' where tl'
1 2
and is are the times during which the vehicle runs at the speeds U I • o:
and Os respectively. Obviously,
Consequently,
18 krn/h
23. The path s travelled by the point in five seconds is equal numerically
to the area enclosed between Oabcd and the time axis (see Fig. 6): 51 = 10.5 em.
The mean velocity of the point in five seconds is VI = SI/t1 = 2.1 cm/~
and the mean acceleration of the point during the same time is
Av
at =7; =0.8 cm/s"
The path travelled in 10 seconds is 52= 25 em.
Therefore, the mean velocity and the mean acceleration are
u 2 = ~2 =2.5 cm/s, a 2 = O.2 em/5 2
2
24. During a small time interval ~t the bow of the boat will move from
point A to point B (Fig. 277). The distance AB=v l ~t, where VI is the speed
of the boat. A rope length of OA-OB=CA=v At will be taken up during
this time. The triangle ABC may be considered as a right one, since AC ~ OA.
v
Therefore VI=--.
COSeL