fisica1-youn-e-freedman-exercicios-resolvidos

3.76: a)
dv
dt
=
=
d
dt
(1/ 2)
vxax
=
v
v
2
x
2
x
d
dt
v
+ v
2
x
( v
y
+ v
2
x
+ v a
b) Using the numbers from Example 3.1 and 3.2,
2
2
dv ( −1.0 m/s)( −0.
50 m/s ) + ( 13 . m/s)(
0.
30 m/s )
=
= 0.54 m/s.
dt
2
2
( −1.0 m/s) + ( 13 . m/s)
The acceleration is due to changing both the magnitude and direction of the velocity. If
the direction of the velocity is changing, the magnitude of the acceleration is larger than
2 2
the rate of change of speed. c)
v ⋅ a
v = v
xax
+ v
ya
y,
v = vx
+ v
y , and so the above form
for dv is seen to be .
dt
v ⋅ a v / v
2
y
2
y
+ v
y
+ v
2
y
.
2
y
)
3.77: a) The path is a cycloid.
v x
b) To find the velocity components, take the derivative of x and y with respect to time:
= Rω ( 1−
cosωt),
and = Rω
sin ωt.
To find the acceleration components, take the
v y
2
2
derivative of v
x
and v
y
with respect to time: a x
= Rω sinωt,
and a y
= Rω cosωt.
c) The particle is at rest ( v v = 0)
every period, namely at t = 0,
2π
/ ω, 4π
/ ω, .... At
y
= x
that time, x = 0,
2πR,
4πR,
...;
and y = 0.
The acceleration is
+ y - direction.
2 2
2
[ ] 1/
Rω
.
2 2 2
d) No, since = ( Rω
sinωt) + ( Rω
cosωt)
a =
2
a = Rω in the
3.78: A direct way to find the angle is to consider the velocity relative to the air and the
velocity relative to the ground as forming two sides of an isosceles triangle. The wind
direction relative to north is half of the included angle, or arcsin( 10/50) = 11. 53° , east of
north.