fisica1-youn-e-freedman-exercicios-resolvidos

3.92:
The x-position of the plane is
( 236 m/s) t and the x-position of the rocket is
2
2
(236 m/s) t + 1/ 2(3.00)(9.80 m/s )cos30°(
t − T ) . The graphs of these two have the form,
If we take y = 0 to be the altitude of the airliner, then
y( t)
T
graph looks like
2
2
2
= −1/
2gT
− gT ( t − T ) + 1/ 2(3.00)(9.80 m/s )(sin 30°
)( t − ) for the rocket. This
By setting y = 0 for the rocket, we can solve for t in terms of
2 2
2
2
2
T,
0 = −(4.90 m/s ) T − (9.80 m/s ) T(
t − T ) + (7.35 m/s )( t − T)
. Using the quadratic
formula for the variable x = t − T, we find
2
(9.80 m/s ) T +
2 2
2 2
(9.80 m/s T ) + (4)(7.35 m/s )(4.9) T
x = t − T =
or t = 2.72T.
Now, using
2
2(7.35 m/s )
the condition that x − x 1000 m,
we find
rocket plane
=
2
2
2
(236 m/s) t + (12.7 m/s ) × ( t − T ) − (236 m/s) t = 1000 m, or (1.72T
)
2 = 78.6 s .
Therefore T = 5.15s.
3.93: a) Taking all units to be in km and h, we have three equations. We know that
heading upstream v
c / w
− vw
/ G
= 2 where v
c / w
is the speed of the curve relative to water
and v
w / G
is the speed of the water relative to the ground. We know that heading
downstream for a time t, ( vc / w
+ vw
/ G
) t = 5.
We also know that for the bottle
v w / G
( t + 1) = 3. Solving these three equations for vw / G
= x, vc
/ w
= 2 + x, therefore
3
( 2 + x + x)
t = 5 or ( 2 + 2x ) t = 5.
Also t = 3/
x −1,
so ( 2 2x )( −1)
= 5 or
+
x
2x 2 + x − 6 = 0. The positive solution is x = v w / G
= 1.5 km/h.
b) v 2 km/h + v 3.5 km/h.
c / w
=
w/G
=