James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 13.4 Motion in Space: Velocity and Acceleration 871
As in the case of one-dimensional motion, the acceleration of the particle is defined as
the derivative of the velocity:
astd − v9std − r0std
y
v(1)
(1, 1)
0
a(1)
x
ExamplE 1 The position vector of an object moving in a plane is given by
rstd − t 3 i 1 t 2 j. Find its velocity, speed, and acceleration when t − 1 and illustrate
geometrically.
SOLUtion The velocity and acceleration at time t are
and the speed is
vstd − r9std − 3t 2 i 1 2t j
astd − r0std − 6t i 1 2 j
FIGURE 2
| vstd | − ss3t 2 d 2 1 s2td 2 − s9t 4 1 4t 2
TEC Visual 13.4 shows animated
velocity and acceleration vectors for
objects moving along various curves.
When t − 1, we have
vs1d − 3 i 1 2 j as1d − 6 i 1 2 j | vs1d | − s13
These velocity and acceleration vectors are shown in Figure 2.
■
Figure 3 shows the path of the par ticle
in Example 2 with the velocity and
acceleration vectors when t − 1.
z
a(1)
v(1)
ExamplE 2 Find the velocity, acceleration, and speed of a particle with position
vector rstd − kt 2 , e t , te t l.
SOLUTION
vstd − r9std − k2t, e t , s1 1 tde t l
astd − v9std − k2, e t , s2 1 tde t l
x
FIGURE 3
1
y
| vstd | − s4t 2 1 e 2t 1 s1 1 td 2 e 2t ■
The vector integrals that were introduced in Section 13.2 can be used to find position
vectors when velocity or acceleration vectors are known, as in the next example.
ExamplE 3 A moving particle starts at an initial position rs0d − k1, 0, 0 l with initial
velocity vs0d − i 2 j 1 k. Its acceleration is astd − 4t i 1 6t j 1 k. Find its velocity
and position at time t.
SOLUtion Since astd − v9std, we have
vstd − y astd dt − y s4t i 1 6t j 1 kd dt
− 2t 2 i 1 3t 2 j 1 t k 1 C
To determine the value of the constant vector C, we use the fact that vs0d − i 2 j 1 k.
The preceding equation gives vs0d − C, so C − i 2 j 1 k and
vstd − 2t 2 i 1 3t 2 j 1 t k 1 i 2 j 1 k
− s2t 2 1 1d i 1 s3t 2 2 1d j 1 st 1 1d k
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