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Math 2204 Multivariable Calculus – Chapter 13: Vector Functions
Sec. 13.4: Motion in Space: Velocity and Acceleration
I.
Velocity and Acceleration
A. Definitions
1. The velocity vector at time t:
r(t + h) − r(t)
!
!
v(t) = r ′(t) = lim
h→0
h
It is also the tangent vector and points in the direction of the tangent line.
2.
The speed of the particle at time
!
t is the magnitude of the velocity vector, v(t) .
ds
!
!
v(t) = r ′(t) =
= rate of change of distance wrt time
dt
3. The acceleration of the particle at time
!
!
!
t is derivative of the velocity: a(t) = v′(t) = r ′′(t) .
B. Examples
1. Find the velocity, and acceleration of a particle with position function
!
!
!
!
r (t) = tet i − sin(3t) j + t k .
2. Given
!
r (t) = 2t 2 − 3t,6t 2 − 7 in the plane, determine the time t when the velocity is
orthogonal to acceleration and then find the speed at that time.
P = (9,5,−2) at time t = 1 is moving with velocity
!
v(t) = 9t 2 , 3t 2 ,2 . Find the position vector.
3. A particle that passes the point
II.
Projectile Motion
A. Newton’s Second Law of Motion
If, at any time t, a force
then
!
!
F(t) = m a(t) .
!
!
F(t) acts on an object of mass m producing an acceleration a(t) ,
B. Parametric Equations of Trajectory
A projectile is fired with angle of elevation
!
α and initial velocity v0 .
Assuming that air
resistance is negligible and the only external force is due to gravity, the parametric
equations of the trajectory are
of the projectile
x = (v0 cos α )t , y = (v0 sin α )t − 12 gt 2 ; v0 is the initial speed
C. Examples
1. A projectile is fired at ground level with an initial speed of
angle of elevation of
287 meters per second at an
61 degrees. How far away will the projectile strike the ground?
4 feet above the ground with an initial speed of 95
feet per second at a launch angle of 42 degrees. How high above the ground will the
projectile be when it is 224 feet downrange?
2. A projectile is fired from a height of
3. A golf ball leaves the ground with an initial speed of
91 feet per second at an angle of
elevation of 30 degrees. What is its maximum height?
III. Tangential and Normal Components of Acceleration
A. Definition
!
!
r (t) be a trajectory with acceleration a(t) . The tangential component of
!
acceleration, aT , is the scalar projection of a onto the unit tangent vector Tˆ .
1. Let
! ! !
!
! ˆ a ⋅ v r ′(t)⋅ r ′′(t)
aT = a ⋅ T = ! =
!
v
r ′(t)
!
aN , is the scalar projection of a onto the unit
!
normal vector N̂ . We could use the formula aN = a ⋅ N̂ to calculate this component.
However, since it can be difficult to calculate N̂ , we’ll make use of the fact that Tˆ and
2. The normal component of acceleration,
N̂ are perpendicular. Below are two methods used to calculate aN .
! !
!
!
axv
r ′(t) x r ′′(t)
aN = ! =
!
v
r ′(t)
or Using the Pythagorean Theorem:
!2
(aT )2 + (aN )2 = a or aN =
!2
a − (aT )2
B. Examples
!
1. A particle moves with the position function r (t) =
t 2 ,t 2t 3 . Find the tangential and
normal components of acceleration.
2. Find the tangential and normal components of acceleration at time
!
−t
t
trajectory is r (t) = e , 2t,e .
t = 2 if the