Section 11.8: Motion in Space: Velocity and Acceleration
In this section, we apply the concepts of tangent and normal vectors to study the motion
of an object along a curve in space.
~
Definition: Suppose a particle moves through space so that its position at time t is R(t).
Then the velocity vector of the particle is
~ 0 (t),
V~ (t) = R
and the speed of the particle at time t is ||V~ (t)||. Similarly, the acceleration of the particle
is given by
~ = V~ 0 (t) = R
~ 00 (t).
A(t)
~
= ht3 , t2 + 1, t3 − 1i, t ≥ 0. Find
Example: The position vector of an object is given by R(t)
its velocity, acceleration, and speed at time t = 1.
The velocity and acceleration at time t are
V~ (t) = h3t2 , 2t, 3t2 i
~
A(t)
= h6t, 2, 6ti.
The speed at time t is
√
√
12t4 + 4t2 .
√
~
At time t = 1, V~ (1) = h3, 2, 3i, A(1)
= h6, 2, 6i, and ||V~ (1)|| = 16 = 4.
||V~ (t)|| =
6t4 + 4t2 + 6t4 =
~
Example: A moving particle starts at initial position R(0)
= h0, 1, 0i with initial velocity
2
~
~
V (0) = h1, 0, 1i. Its acceleration is A(t) = ht, t , cos 2ti. Find its velocity and position at
time t.
~ = V~ 0 (t),
Since A(t)
V~ (t) =
Z
Z
=
=
~
A(t)dt
ht, t2 , cos 2tidt
1 2 1 3 1
~
t , t , sin 2t + C.
2 3 2
~ = h1, 0, 1i. Then
Using the initial velocity, V~ (0) = C
1
1
1
2
3
V~ (t) =
t + 1, t , sin(2t) + 1 .
2
3 2
~ 0 (t),
Since V~ (t) = R
~
R(t)
=
Z
V~ (t)dt
Z 1 2
1 3 1
=
t + 1, t , sin(2t) + 1 dt
2
3 2
1 3
1 4 1
~
=
t + t, t , − cos(2t) + t + C.
6
12
4
Using the initial position,
1
~ = h0, 1, 0i.
~
+C
R(0)
= 0, 0, −
4
~ = h0, 1, 1/4i and
Thus, C
~
R(t)
=
1 3
1
1
1
t + t, t4 + 1 − cos(2t) + t +
6
12
4
4
.
Example: A force of 20 N acts directly upward from the xy-plane on an object with mass 4
kg. The object starts at the origin with initial velocity V~ (0) = h1, −1, 0i. Find its position
function and its speed at time t.
By Newton’s Second Law of Motion,
~
F~ (t) = mA(t)
~
h0, 0, 20i = 4A(t)
~
A(t)
= h0, 0, 5i.
~ = V~ 0 (t),
Since A(t)
V~ (t) =
Z
~
A(t)dt
=
Z
~
h0, 0, 5idt = h0, 0, 5ti + C.
~ = h1, −1, 0i. Thus,
Using the initial velocity, V~ (0) = C
V~ (t) = h1, −1, 5ti.
Therefore, the speed of the object is
||V~ (t)|| =
√
2 + 25t2 .
~ 0 (t),
Since V~ (t) = R
~
R(t)
=
Z
V~ (t)dt =
Z
h1, −1, 5tidt =
5
t, −t, t2
2
~
+ C.
~
~ = ~0. Thus,
Using the initial position, R(0)
=C
5
2
~
R(t)
= t, −t, t .
2
Note: When studying the motion of a particle in space, it is often useful to break the
acceleration down into two components: one in the direction of the unit tangent and one
in the direction of the unit normal. These are known as the tangential and normal
components of acceleration.
Figure 1: Illustration of the tangential and normal components of acceleration.
Theorem: (Tangential and Normal Components of Acceleration)
~
Suppose that the position of a particle at time t is given by R(t).
Then the tangential and
normal components of the particle’s acceleration are
AT =
~ 0 (t) · R
~ 00 (t)
R
~ 0 (t)||
||R
and
AN =
~ 0 (t) × R
~ 00 (t)||
||R
.
~ 0 (t)||
||R
~
Example: A particle moves with position function R(t)
= ht3 , t2 , ti. Find the tangential and
normal components of acceleration at a general position and at the origin.
First,
~ 0 (t) = h3t2 , 2t, 1i
R
~ 00 (t) = h6t, 2, 0i
R
√
~ 0 (t)|| = 9t4 + 4t2 + 1
||R
~ 0 (t) · R
~ 00 (t) = 18t3 + 4t
R
~i ~j ~k ~ 0 (t) × R
~ 00 (t) = 3t2 2t 1 = h−2, 6t, −6t2 i
R
6t 2 0 √
√
~ 0 (t) × R
~ 00 (t)|| = 4 + 36t2 + 36t4 = 2 1 + 9t2 + 9t4 .
||R
Then
18t3 + 4t
AT = √
9t4 + 4t2 + 1
At the origin (t = 0), AT = 0 and AN = 2.
and
√
2 1 + 9t2 + 9t4
AN = √
.
9t4 + 4t2 + 1