Transcript 12.3 Velocity and Acceleration

12
Vector-Valued Functions
Copyright © Cengage Learning. All rights reserved.
HWQ 3/19/15
Find v(t) given the following conditions:
v  t   4cos ti  3j
v   0   3j
v  0  4i

3t 2 
v  t    4cos t  i +  3t 
j
2 

12.3
Day 1
Velocity and Acceleration
Copyright © Cengage Learning. All rights reserved.
Objectives
 Describe the velocity and acceleration associated
with a vector-valued function.
 Use a vector-valued function to analyze projectile
motion.
Velocity and Acceleration
As an object moves along a curve in the plane, the coordinates x
and y of its center of mass are each functions of time t.
Rather than using the letters f and g to represent these two
functions, it is convenient to write x = x(t) and y = y(t).
So, the position vector r(t) takes the form
r(t) = x(t)i + y(t)j.
Velocity and Acceleration
Velocity and Acceleration
The velocity vector is the tangent vector to
the curve at point P
Speed  v  t 
“Absolute value” means “distance
from the origin” so we must use the
Pythagorean theorem.
The magnitude of the velocity vector r'(t) gives the
speed of the object at time t.
velocity vector
Direction 
speed

v t 
v t 
Velocity and Acceleration
For motion along a space curve, the definitions are similar.
That is, if r(t) = x(t)i + y(t)j + z(t)k, you have
Velocity = v(t) = r'(t) = x'(t)i + y'(t)j + z'(t)k
Acceleration = a(t) = r''(t) = x''(t)i + y''(t)j + z''(t)k
Speed =
Example 1 – Finding Velocity and Acceleration Along a Plane
Curve
Find the velocity vector, speed, and acceleration vector of a
particle that moves along the plane curve C described by
Solution:
The velocity vector is
The speed (at any time) is
Example 1 – Solution
The acceleration vector is
Note that the velocity and acceleration
vectors are orthogonal at any point in
time. This is characteristic of motion
at a constant speed.
Example 1, con’t
Write the parametric and rectangular
equations for r(t).
Because the velocity vector
has a constant magnitude,
but a changing direction as t increases, the particle moves
around the circle at a constant speed..
Example 2 (you try):
r  t    3cos t  i   3sin t  j
a) Find the velocity and acceleration vectors.
dr
v
  3sin t  i   3cos t  j
dt
dv
a
  3cos t  i   3sin t  j
dt
b) Find the velocity, acceleration, speed and direction
of motion at t   / 4 .

Example 2:
r  t    3cos t  i   3sin t  j
dr
v
  3sin t  i   3cos t  j
dt
dv
a
  3cos t  i   3sin t  j
dt
b) Find the velocity, acceleration, speed and direction
of motion at t   / 4 .
3
3
 

  
i
j
velocity: v     3sin  i   3cos  j  
4 
4
2
2
4 
 

3
3
  
i
j
acceleration: a     3cos  i   3sin  j  
4 
4
4 
2
2

Example 2:
r  t    3cos t  i   3sin t  j
dr
v
  3sin t  i   3cos t  j
dt
dv
a
  3cos t  i   3sin t  j
dt
b) Find the velocity, acceleration, speed and direction
of motion at t   / 4 .
3
3
 
v   
i
j
2
2
4
3
3
 
a   
i
j
2
2
4
2
speed:
direction:
    3   3 
v 

 

2  2

4
v  / 4 
v  / 4 
2
9 9


3
2 2
3 / 2
3/ 2   1 i  1 j

i
j
2
2
3
3

Example 3:
 

r  t   2t 3  3t 2 i  t 3  12t j
a) Write the equation of the tangent where

 
t  1.

dr
v t  
 6t 2  6t i  3t 2  12 j
dt
At t  1 :
position:
tangent:
v  1  12i  9 j
r  1  5i  11j
 5,11
slope:
y  y1  m  x  x1 
3
y  11    x  5 
4
9
3

12
4
3
29
y  x
4
4

Example 3:

 

r  t   2t 3  3t 2 i  t 3  12t j

 

dr
v t  
 6t 2  6t i  3t 2  12 j
dt
b) Find the coordinates of each point on the path where
the horizontal component of the velocity is 0.
2
6
t
 6t .
The horizontal component of the velocity is
6t  6t  0
2
t t  0
r  0   0i  0 j
 0, 0 
2
t  t  1  0
t  0, 1
r 1   2  3 i  1 12  j
r 1  1i  11j
 1, 11
 0, 0 
 1, 11
Example 4:
An object starts from rest
at the point (1,2,0) and
moves with an acceleration
of a  t   j + 2k ,
where a  t  is measured
in feet per second per
second. Find the location
of the object after 2
seconds.
 t2

r t   i    2  j + t 2k
2

r  2  1  4 j + 4k
Example 5 (you try) :
Use the given acceleration function to find the
velocity and position vectors. Then find the
position at time t = 2.
a  t    cos t i - sint j
v  0  j + k
r  0 i
r  2   cos2 i  sin 2 j + 2k
Homework:
Section 12.3
Day 1: pg. 854 # 1-23 odd
Day 2: MMM pg. 181 +
12.3, pg. 854 #25, 34, 37, 38
Day 2: Projectile Motion
Fort Pulaski, GA
Photo by Vickie Kelly, 2002
Greg Kelly, Hanford High School, Richland, Washington
One early use of calculus was to study projectile motion.
In this section we assume ideal projectile motion:
Constant force of gravity in a downward direction
Flat surface
No air resistance (usually)

vo

We assume that the projectile is launched from the origin
at time t =0 with initial velocity vo.
Let vo  vo
then vo   vo cos  i   vo sin   j
The initial position is:
ro  0i  0 j  0

vo

Newton’s second law
of motion:
f  ma
2
d r
f m 2
dt
Vertical acceleration

vo

Newton’s second law
of motion:
f  ma
2
d r
f m 2
dt
The force of gravity is:
f   mg j
Force is in the downward direction

vo

Newton’s second law
of motion:
f  ma
2
d r
f m 2
dt
The force of gravity is:
f   mg j
2
d r
m 2  mg j
dt

vo

Newton’s second law
of motion:
f  ma
2
d r
f m 2
dt
The force of gravity is:
f   mg j
2
d r
m 2  mg j
dt

2
d r
 g j
2
dt
Initial conditions: r  ro
dr
 vo
dt
when t  o
dr
  gt j  v o
dt
1 2
r   gt j  v ot  ro
2
1 2
r   gt j   vo cos  t i   vo sin   t j 0
2

1 2
r   gt j   vo cos   t i   vo sin   t j  0
2
Vector equation for ideal projectile motion:
1 2

r   vo cos   t i    vo sin   t  gt  j
2


1 2
r   gt j   vo cos   t i   vo sin   t j  0
2
Vector equation for ideal projectile motion:
1 2

r   vo cos   t i    vo sin   t  gt  j
2


Parametric equations for ideal projectile motion:
x   vo cos   t
1 2
y   vo sin   t  gt
2

Example 1:
A projectile is fired at 60o and 500 m/sec.
Where will it be 10 seconds later?
x  500  cos 60 10
x  2500
1
y  500sin 60 10   9.8 10 2
2
y  3840.13
The projectile will be 2.5 kilometers downrange and
at an altitude of 3.84 kilometers.
Note: The speed of sound is 331.29 meters/sec
Or 741.1 miles/hr at sea level.

The maximum height of a projectile occurs when
the vertical velocity equals zero.
dy
 vo sin   gt  0
dt
vo sin   gt
vo sin 
t
g
time at maximum height

The maximum height of a projectile occurs when
the vertical velocity equals zero.
dy
 vo sin   gt  0
dt
vo sin   gt
vo sin 
t
g
We can substitute this expression into the formula
for height to get the maximum height.

1 2
y   vo sin   t  gt
2
ymax
vo sin  1  vo sin  
  vo sin  
 g

g
2  g 
ymax 
 vo sin  
g
2
vo sin  


2
2
2g

1 2
y   vo sin   t  gt
2
ymax
vo sin  1  vo sin  
  vo sin  
 g

g
2  g 
2  vo sin    vo sin  


2g
2g
2
ymax
ymax
vo sin  


2g
2
2
2
maximum
height
(not including initial height)

When the height is zero:
time at launch:
1 2
0   vo sin   t  gt
2
1 

0  t  vo sin   gt 
2 

t 0

When the height is zero:
time at launch:
t 0
1 2
0   vo sin   t  gt
2
1 

0  t  vo sin   gt 
2 

1
vo sin   gt  0
2
1
vo sin   gt
2
time at impact
(flight time)
2vo sin 
t
g

If we take the expression for flight time and substitute
it into the equation for x, we can find the range.
x  vo cos  t
2vo sin 
x  vo cos  
g
If we take the expression for flight time and substitute
it into the equation for x, we can find the range.
x  vo cos  t
2vo sin 
x  vo cos  
g
2
vo
x
 2cos sin  
g
2
vo
x
sin  2 
g
Range

The range is maximum when
sin  2 
is maximum.
sin  2   1
2  90o
  45
o
Range is maximum
when the launch
angle is 45o.
2
vo
x
sin  2 
g
Range

If we start somewhere besides the origin, the equations
become:
x  xo   vo cos  t
1 2
y  yo   vo sin   t  gt
2

Example 4:
A baseball is hit from 3 feet above the ground with an
initial velocity of 152 ft/sec at an angle of 20o from the
horizontal. A gust of wind adds a component of -8.8 ft/sec
in the horizontal direction to the initial velocity.
The parametric equations become:
x  152 cos 20  8.8  t
o
vo

y  3  152sin 20o  t  16t 2
yo
1
g
2

These equations can be graphed on the TI-89 to model
the path of the ball:
t2
Note that the calculator is in degrees.


Using
the
trace
function:
Max height about 45 ft
Time
about
3.3 sec
Distance traveled about 442 ft

Example 6 – Describing the Path of a Baseball
A baseball is hit 3 feet above ground level at
100 feet per second and at an angle of 45° with respect
to the ground, as shown in Figure 12.19. Find the
maximum height reached by the baseball. Will it clear a
10-foot-high fence located 300 feet from home plate?
x   vo cos   t
1 2
y   vo sin   t  gt
2
Example 6 – Solution
The maximum height occurs when
which implies that
So, the maximum height reached by the ball is
Example 6 – Solution
The ball is 300 feet from where it was hit when
Solving this equation for t produces
At this time, the height of the ball is
= 303 – 288
= 15 feet.
Therefore, the ball clears the 10-foot fence for a home run.
Homework:
Section 12.3
Day 1: pg. 854 # 1-23 odd
Day 2: MMM pg. 181 + 12.3, pg. 854
#25, 34, 37, 38