fisica1-youn-e-freedman-exercicios-resolvidos

2.64: Taking the start of the race as the origin, runner A's speed at the end of 30 m can
be found from:
v
2
A
= v
2
0A
+ 2a
v
A
A
=
( x − x
0
96 m
2
2
) = 0 + 2(1.6 m s )(30 m) = 96 m
2
s
2
= 9.80 m s
s
2
A’s time to cover the first 30 m is thus:
vA − v0A
9.80 m s
t = = = 6.13s
2
aA
1.6 m s
and A’s total time for the race is:
(350 − 30) m
6 . 13s +
= 38.8 s
9.80 m s
B’s speed at the end of 30 m is found from:
v
2
B
= v
2
0B
+ 2a
B
( x − x
0
2
) = 0 + 2(2.0 m s )(30 m) = 120 m
2 2
vB
= 120 m s = 10.95 m s
B’s time for the first 30 m is thus
vB − v0B
10.95 m s
t = = = 5.48 s
2
aB
2.0 m s
and B's total time for the race is:
(350 − 30) m
5 . 48s +
= 34.7 s
10.95 m s
2
s
2
B can thus nap for
38 .8 − 34.7 = 4.1s
and still finish at the same time as A.
2.65: For the first 5.0 s of the motion, = 0, t 5.0 s.
v0 x
=
vx = v 0 x
+ axt
gives v
x
= a x
(5.0 s).
This is the initial speed for the second 5.0 s of the motion. For the second 5.0 s:
v = a 5.0 s), t = 5.0 s, x − x 150 m.
0 x x
(
0
=
1 2
2
2
2
− x0 = v0xt
+ a gives 150 m = (25 s ) + (12.5 s ) and = 4.0 m s
2 xt
a
x
ax
a
x
x
Use this a x and consider the first 5.0 s of the motion:
1 2 1
2
2
x − x = v t + a t = 0 + (4.0 m s )(5.0 s) 50.0 m.
0 0x
2 x
2
=