V t p t x 8 sin4 t given 8 cos4 t given y a t p t x

v ( t ) = p
¤
( t ) =
X
-8 sin(4 t )
given
, 8 cos(4 t )
given
Y
a ( t ) = p
¤¤
( t ) =
X
-32
2
cos(4 t )
given
, -32
2
sin(4 t )
given
Y
For the physical interpretation, recall the classic physics
equation, “Force = mass × acceleration.” A force acting
on a mass induces acceleration (i.e., the mass moves);
acceleration acting on a mass induces a force (gravity
gives our mass a weight ). Thus force and acceleration
are closely related. A moving ball “wants” to travel in
a straight line. Why does the ball in our example move
in a circle? It is attached to your hand by a string. The
string applies a force to the ball, affecting it’s motion:
the string accelerates the ball. This is not acceleration
in the sense of “it travels faster;” rather, this acceleration
is changing the velocity of the ball. In what direction is
this force/acceleration being applied? In the direction of
the string, towards your hand.
Hence it makes sense a ( t ) is parallel to p ( t ) , but has a
different magnitude and points in the opposite direction.
The magnitude of the acceleration is related to the speed
at which the ball is traveling. A ball whirling quickly
is rapidly changing direction/velocity. When velocity is
changing rapidly, the acceleration must be large .
Question 164. An object moves in a spiral with position func-
tion
p ( t ) = OE cos t, sin t, t º ,
where distances are measured in meters and time is in minutes.
Describe the object’s velocity and acceleration at time t .
v ( t ) =
(
- sin t , cos t , 1
)
a ( t ) =
(
- cos t , - sin t , 0
)
.
Question 165. What is the speed of this object?
v ( t ) =
ø
2 m_ min
Question 166. What is the angle between v and a ?
The angle equals _2 radians
Feedback( correct ): Since the speed is constant, the velocity and
the acceleration are perpendicular.
Projectile motion
An important application of vector-valued position functions is
projectile motion : the motion of objects under the influence of
gravity. We will measure time in seconds, and distances will
either be in meters or feet. We will show that we can completely
describe the path of such an object knowing its initial position
and initial velocity (where it is and where it is going. )
Suppose an object has initial position
p (0) = OE x (0) , y (0) º
300

Motion and paths in space
and initial velocity of
v (0) = v
0
OE cos( ) , sin( ) º .
Here, is often called the angle of elevation . Since the accel-
eration of the object is known, namely
a ( t ) = OE 0 , - g º ,
where g is the gravitational constant, we can find p ( t ) knowing
our two initial conditions. We first find v ( t ) :
v ( t ) =
˚
a ( t ) dt
v ( t ) =
˚
OE 0 , - g º dt
v ( t ) = OE 0 , - gt º + v (0) .
We integrate once more to find p ( t ) :
p ( t ) =
˚
v ( t ) dt
p ( t ) =
˚
OE 0 , - gt º + C
1
dt
p ( t ) =
@
0 ,
- gt
2
2
A
+ t v (0) + p (0) .
You can adjust the initial position, P
0
, angle, magnitude of the
velocity, and magnitude of the acceleration below:
Geogebra link:
We demonstrate how to solve for a position function in the
context of projectile motion in the next example.