Calculus by Briggs and Cochran
Section 3.5- Derivatives as Rates of Change
9.
A patrol car leaves the station at 9:00 A.M. heading north with
position function    ab that gives its location in miles  hours
9:00 A.M. (see figure). Assume  is positive when the car is
north of the patrol station.
a)
Determine the average velocity of the car during the 1st
45 minutes of the trip.
45 minutes is 3/4   hr. From    to 0.75, 
changes from 0 to 30 which means the average velocity

 

   mph.
is

 
b)
Find the average velocity of the car over the interval
c d Is the average velocity a good estimate of the
velocity at 9:30 A.M.?
From    to     goes changes from 10 to 30
  

which means the average velocity is

  

  mph.
At 9:00,    For the interval c d, the midpoint,
0.50, gives a very good approximation.
c)
Find the average velocity of the car over the interval
c d Estimate the velocity of the car at 11:00 A.M.
and determine the direction in which the patrol car is
moving.
On the interval c d,  goes from  to  
Therefore the average velocity is
   ab
 

   mph. What is happening is
  

the car has reversed direction (from north to south) when
  1.1 which is a little past 10:00 A.M..
  mph is a good approximation of the velocity at 11:00
because 11:00 means    which is the midpoint of
the interval c d. The negative velocity means the
car is heading south.
d)
Describe the motion of the patrol car relative to the patrol
station between 9:00 and 12:00.
Note that when the slope of the curve is positive, the car
is heading north, and when the slope of the car is negative,
the car is heading south. Also, when    , the car is
at the station (or passing the station).
When the slope of the tangent line is positive (   to 1.1),
the car is heading north away from the station. At a little
past 1.1, the car reverses direction and heads south until
   when the slope changes to positive. Note that when
   at 11:00, the car is driving past the station again, but
is heading south. When    hrs   hrs,  minutes 
1:40 P.M., the car is heading north again.
15.
Position, velocity, and acceleration- Suppose the position of an
object moving horizontally after  seconds is given by the position
function    ab           
where  is measured in feet.
a) Graph the position function.
50
40
30
Position function
f(t)
20
10
1
b)
2
3
4
5
6
Find and graph the velocity function. When is the object
stationary, moving to the right, and moving to the left?
        a    b  a  ba  b
   when    and . Mark off  and  on a number
line and make test values. For example, when   
  abab    moving to the right.
When      aba  b    moving to left.
When      a  ba  b    moving to right.
----------------- 2 ------------- 5 ---------------To right
To left
To right
The object is stationary when    sec and  sec.
The object is moving to the right when     
and also when   
The object is moving to the left when     
The graph of ab       is below.
60
50
40
30
20
10
- 10
c)
d)
v = f(t)
1
2
3
4
5
6
Determine the velocity and acceleration of the object at
  

 ab
       ab      

  ft/sec.


     ab        ft/sec .

Determine the acceleration of the object when its
velocity is 0.
The velocity is 0 when    and    Therefore
ab  ab          ft/sec and
ab  ab         ft/sec .
17.
A stone is thrown vertically upward from the edge of a cliff
with an initial velocity of 64 ft/sec from a height of 32 ft above
the ground. The height  (in ft) of the stone above the ground
 seconds after it is thrown is        
a) Determine the velocity  of the stone after  seconds.


     

b) When does the stone reach its highest point?
The stone reaches its highest point when   
             
Therefore the stone reaches its highest point
after 2 seconds.
c) What is the height of the stone at the highest point?
The highest point is ab   ab  ab     ft.
d) When does the stone strike the ground?
The stone hits the ground when    
               
 a    b           
  È  aba  b
  È





  È
We don't want the negative, so  
  sec.

e) With what velocity does the stone strike the ground?
The stone hits the ground after  seconds. The
velocity when it hits the ground is
ab   ab      ft/sec.