fisica1-youn-e-freedman-exercicios-resolvidos

2.59: (a) Denote the time for the acceleration (4.0 s) as t 1 and the time spent running at
constant speed (5.1 s) as t 2 . The constant speed is then at 1 , where a is the unknown
acceleration. The total l is then given in terms of a, t 1 and t 2 by
and solving for a gives
1 2
l = at + at
2
1 1t2
,
l
a =
2
(1 2) t + t t
1
1 2
(100 m)
=
2
(1 2)(4.0 s) + (4.0 s)(5.1s)
= 3.5 m
s
2
.
(b) During the 5.1 s interval, the runner is not accelerating, so a = 0.
2
2
(c) ∆v
∆t
= [(3.5 m s )(4 s)] (9.1s) = 1.54 m s .
(d) The runner was moving at constant velocity for the last 5.1 s.
2.60: a) Simple subtraction and division gives average speeds during the 2-second
intervals as 5.6, 7.2 and 8.8 m s .
b) The average speed increased by 1.6 m s during each 2-second interval, so the
acceleration is 0.8
2
m s .
c) From Eq. (2.13), with v 0 = 0, v = 2(0.8 m s )(14.4 m) = 4.8 m s.
Or,
recognizing that for constant acceleration the average speed of 5.6 m/s is the speed one
2
second after passing the 14.4-m mark, 5.6 m s – (0.8 m s )(1.0 s) = 4.8 m s .
d) With both the acceleration and the speed at the 14.4-m known, either Eq. (2.8)
or Eq. (2.12) gives the time as 6.0 s.
2
e) From Eq. (2.12), x – x 0 = (4.8 m s )(1.0 s) + 21
(0.8 m s )(1.0 s) 2 = 5.2 m. This
is also the average velocity (1/2)(5.6 m s + 4.8 m s ) times the time interval of
1.0 s.
2
2.61: If the driver steps on the gas, the car will travel
2
2
(20 m s)(3.
0 s) + (1 2)(2.3 m s )(3.0 s) = 70.4m.
If the brake is applied, the car will travel
2
2
(20 m s)(3.
0 s) + (1 2)( −3.8
m s )(3.0 s) = 42.9 m,
so the driver should apply the brake.