fisica1-youn-e-freedman-exercicios-resolvidos

3.45: a) The a = 0 and a y
= −2β
, so the velocity and the acceleration will be
x
perpendicular only when v = 0 , which occurs at t = 0 .
y
2 2 2 2
b) The speed is v = ( α + 4β
t )
1/ , dv / dt = 0 at t = 0 . (See part d below.)
c) r and v are perpendicular when their dot product is 0:
2
2
2 3
( αt )( α)
+ (15.0 m − βt ) × ( −2βt)
= α t − (30.0 m) βt + 2β
t = 0 . Solve this for t:
2
(30.0 m)(0.500 m/s ) −(1.2 m/s)
t = ±
= + 5.208 s , and 0 s, at which times the student is at (6.25 m,
2 2
2(0.500 m/s )
2
1.44 m) and (0 m, 15.0 m), respectively.
d) At t = 5.208 s , the student is 6.41 m from the origin, at an angle of 13 ° from the x-
2 2 1/ 2
axis. A plot of d ( t)
= ( x(
t)
+ y(
t)
) shows the minimum distance of 6.41 m at 5.208
s:
e) In the x - y plane the student’s path is:
β 3 γ 2
3.46: a) Integrating, r = ( α t − t )ˆ i + ( t ) ˆ
r
j . Differentiating, a = ( −2β )ˆ i + γ ˆj
.
3
2
b) The positive time at which x = 0 is given by t 2 = 3α β . At this time, the y-coordinate
is
2
γ 2 3αγ
3(2.4 m/s)(4.0 m/s )
y = t = =
= 9.0 m
3
2 2β
2(1.6 m/s )
.