fisica1-youn-e-freedman-exercicios-resolvidos

2.14: (a) The displacement vector is:
r
( t)
= −(5.0
m s) tiˆ
+ (10.0 m s) tˆj
+ ((7.0
m s) t − (3.0 m s
2 ) t 2 )kˆ
The velocity vector is the time derivative of the displacement vector:
r
d
( t)
= ( −5.0
m s)ˆ i + (10.0 m s) ˆj
+ (7.0 m s − 2(3.0 m s
2 ) t ) kˆ
dt
and the acceleration vector is the time derivative of the velocity vector:
At t = 5.0 s:
2
d r t)
= −6.0
m s
2
dt
( 2
kˆ
r
i
ˆ
2
2
( t)
= −(5.0
m s)(5.
0 s)ˆ + (10.0 m s)(5.0 s) j + ((7.0m s)(5.0 s) − ( −3.0
m s )(25.0 s ))kˆ
= ( −25.0 m)ˆ i + (50.0 m) ˆj
− (40.0 m) kˆ
d r ( t)
i
ˆ
2
= ( −5.0
m s)ˆ + (10.0 m s) j + ((7.0 m s − (6.0 m s )(5.0 s)) kˆ
dt
= ( −5.0
m s)ˆ i + (10.0 m s) ˆj
− (23.0 m s) kˆ
2
d r ( t)
2
= −6.0
m s kˆ
2
dt
(b) The velocity in both the x- and the y-directions is constant and nonzero; thus
the overall velocity can never be zero.
(c) The object's acceleration is constant, since t does not appear in the acceleration
vector.