Velocity and Acceleration
Suppose that an object moves along a curve traced out by the endpoint
of the vector valued function rab  a ab abb   ab  ab
Then  ab is a vector pointing in the direction of motion and the
magnitude (speed) of this object is given by l abl
Def: The derivative of ab is called the velocity vector.
 ab  ab is the velocity and l abl is the speed.
The derivative of the velocity vector ab is the
acceleration vector denoted by ab
ab   ab   ab
The vector ab gives information about how the velocity is
changing. Note that the acceleration can change the velocity in
2 ways- the direction can be changing as well as the speed.
For example, we could be going around a circle at a constant speed.
Then the acceleration vector is perpendicular to the direction of motion.
Thus the acceleration will not change the speed in any way. This will
be illustrated in the 2nd example that follows. On the other hand if the
acceleration vector is parallel to the velocity vector, then the speed will
change, but the direction will not.
Example:
Find the velocity and acceleration vectors if the position of an object is
given by ab  a   b where    Sketch the
path of the object. Draw the velocity and acceleration vector when   
 ab  a  b and ab  a b
 ab  a b     and ab  a b   
Speed  È    that speed is increasing as t increases.
As time goes on, the direction changes less and less (why?).
v = 2i + j
y
(1,1)
a = 2i
x
The acceleration is a constant 
Example
Say the motion of a particle is given by a   b This motion is
circular. Note that the speed is constant, but the direction is always
changing. Therefore the ab  
ab  a     b
 ll  speed  È       
ab  a      b 
ab  a b ab  a b    ab  a   b   
y
v=j
(1,0)
x a =-i
Note that the velocity vector is tangent to the curve. At
a b the motion is upward. But the acceleration is
directed toward the origin. ab is always tangent to
the curve, and ab is always pointing to the center when
the motion is circular. Note that in this example,  and  are
always orthogonal. This is not always the case.
Theorem: If the speed is constant then the acceleration vector is always
perpendicular to the velocity vector.
Proof:
speed    ll         

a  b                    

       the angle between  and  is 90°.
We did this proof in the previous section.
Example:
Find the velocity and position of an object at any time  , given that its
acceleration is ab  a     b its initial vector is
ab  a  b and its initial position is ab  a  b
ab   ab 
ab  ( ab  ( ˆ        ‰ 
   a  b     a    b.
Note: it's easier to write      as a   b
ab  a  b  a   b  a    b  a  b 
                    
Therefore ab  a      b  a  b 
a       b 
ab  ( ˆ       ‰  
a        b  a   b 
a             b
ab  a     b  a  b          
Therefore ab  a          b
Example:
A baseball is hit from a height of 3 feet with initial speed of 120 feet per
second and at an angle of 30° above the horizontal. find a vector valued
function describing the position of the ball  seconds after it is hit. To be
a home run, the ball must clear a wall 385 feet away and 6 feet tall.
Determine whether this is a home run.
ht = 6 ft
120
30
385 ft
ab      °     °  a b
The acceleration due to gravity is 32 ft/sec  ab  a  b
ab  ( a  b  a  b  a b  a    b
where a b is the initial velocity.
ab  a b          
and ab  a   b
ab  ( a   b  ˆ       ‰
where now a b is the initial position which i a b ,
and ab  a b  a b 
ab  a      b
To see if it's a home run, see when the i-component is 385.

     
  seconds 

The j-component       must be larger than 6 if it's a home
run.
 ab  ab     ft
It just misses being a home run