fisica1-youn-e-freedman-exercicios-resolvidos

.0 m
2.56: a) 1.25m s.
25
=
20.0 s
25 m
=
15 s
b) 1.67m s.
c) Her net displacement is zero, so the average velocity has zero magnitude.
50 .0 m
d) = 1.43 m s.
Note that the answer to part (d) is the harmonic mean, not the
35.0 s
arithmetic mean, of the answers to parts (a) and (b). (See Exercise 2.5).
2.57: Denote the times, speeds and lengths of the two parts of the trip as t 1 and t 2 , v 1 and
v 2 , and l 1 and l 2.
a) The average speed for the whole trip is
l
t
1
+ l
+ t
2
l1
+ l2
(76 km) + (34 km)
=
=
76 km
( l ) ( ) ( ) ( )
34 km
1
v1
+ l2
v2
88 km h
+
72 km h
1 2
=
82 km
h,
or 82.3 km/h, keeping an extra significant figure.
b) Assuming nearly straight-line motion (a common feature of Nebraska
highways), the total distance traveled is l 1 –l 2 and
v
l
− l
(76 km) − (34 km)
=
76 km 34 km
( ) + ( )
1 2
ave
=
=
t1
+ t2
88 km h 72 km h
31 km
h.
( 31.4
km hr to three significant figures.)
2.58: a) The space per vehicle is the speed divided by the frequency with which the cars
pass a given point;
96 km h
= 40 m vehicle.
2400 vehicles h
An average vehicle is given to be 4.5 m long, so the average spacing is 40.0 m – 4.6 m
= 35.4 m.
b) An average spacing of 9.2 m gives a space per vehicle of 13.8 m, and the traffic
flow rate is
96000 m h
= 6960 vehicle h.
13.8 m vehicle