*Chapter
20 - 4 Velocity and Acceleration*
*Derivatives can be related
very easily to physics applications. We can relate it to the position
function, usually denoted as s(t) or h(t), the velocity function denoted
v(t), and the acceleration function denoted a(t). Notice that are
a function of time! Warning!!
Warning!! Make sure you understand the
difference between average and instantaneous. The *__average
velocity__ can be described as the change
between two points, thus giving you the __slope of the line connecting
these two points__! While the __instantaneous
velocity__ gives you the slope at a single
point in time, thus giving you the __slope of the tangent line__!
That's right!! It's the !!!
*Average formulas*

*Where "s" is the position
at any time "t"*
* *
*Instantaneous Formulas*
*v(t) = s'(t)
a(t) = v'(t) = s"(t)*
*Where v(t)
is the first derivative of the position function
and a(t) is the
first derivative of the velocity function.
Also note it is the
derivative of the position function!!*
* *

* *
*Comments*
*v(t) = 0 means the object
is not moving.*
*v(t) = 3 means the object
would be moving forward (positive)*
*v(t) = -2 means the object
would be moving backward (negative)*
*a(t) = 0 means there
is no change in the velocity*
*a(t) = 5 means the object
is going faster (positive)*
*a(t) = -3 means the object
is slowing down.*
*Sample problems*
*1) s(t) = t*^{2}
- 20. find the average velocity from t = 3 to t = 5.
*Solution: Use the average
velocity formula!*

* *

*Therefore, the average velocity
is *__8__

*2) Find the velocity function
and the acceleration function for the function s(t) = 2t*^{3}
+ 5t - 7
*Solution: Use the instantaneous
formulas*

*v(t) = s'(t) = 6t*^{2}
+ 5
*a(t) = v'(t) = 12t*
*b) Find the velocity
and acceleration at t = 2 for the above function.*
*Solution: Simply replace
t with 2 in the above formulas*

*v(2) = 6(4) + 5 = 29*
*a(2) = 12(2) = 24*
* *
*3) If a ball is thrown
vertically upward with an initial velocity of 128 ft/sec, the ball's height
after t seconds is s(t) = 128t - 16t*^{2}
*a) What is the velocity
function?*
*Solution: v(t) = 128
- 32t*

*b) What is the velocity
when t = 2, 4, 6*
*Solution: v(2) = 128
- 32(2) = 64*
* v(4) =
128 - 32(4) = 32*
* v(6) =
128 - 32(6) = -64*

*c) At what time is the
velocity 48 ft/sec? 16 ft/sec? -48 ft/sec?*
*Solution: Set the velocity
function to each of the above values!*

*48 = 128 - 32t
16 = 128 - 32t -48
= 128 - 32t*
*
-80 = -32t
-112 = -32t
-176 = -32t*
*
2.5 = t
3.5 = t
5.5 = t*

*d) When is the velocity
zero?*
*Solution: Set the velocity
function to zero and solve*

*0 = 128 - 32t*
*-128 = -32t*
*4 = t*
*e) What is the height
of the ball at the time the velocity is 0?*
*Solution: let t = 4
in the position function*

*s(4) = 128(4) - 16(16)
= 512 - 256 = 256'*
*f) Is the answer to
part "e" the maximum height?*
*Solution: Yes, because
it is the zero of the first derivative.*

*g) When does the ball hit
the ground?*
*Solution: Set the position
function to zero and solve!*

*128t - 16t*^{2}
= 0
*16t(8 - t) = 0*
*At t = 0, or t = 8.
Think. At t = 0, you haven't thrown the ball yet! The ball
will hit the ground in 8 seconds. You could have reasoned that it
took 4 seconds to reach maximum height and another 4 secs to come back
down.*
*h) What is the velocity
when the ball hits the ground?*
*Solution: Find v(8)
v(8) = 128 - 32(8) = -128 ft/sec. It is negative because the ball
is coming back down! Notice it has the same velocity as when it was
thrown upward!*

*i) What is the acceleration
function?*
*Solution: Find a(t)
by finding the derivative of v(t)!*

*a(t) = -32 ft/sec*^{2}
*j) What is the acceleration
at t = 3, 5, 8?*
*Solution: a(3) = -32
a(5) = -32 a(8) =
-32*

*No matter what time it
will always be -32. (This is the effect of gravity.)*
*It's now time to get
ready for the big test!!*
*Sample test is next so
have plenty of water (notice the motif) and hang on to your hat!*