3.7 RATES OF CHANGE IN THE
NATURAL AND SOCIAL SCIENCES
RELATIONSHIP BETWEEN POSITION,
VELOCITY, AND ACCELERATION
• Velocity is the first derivative of position
• Acceleration is the first derivative of velocity
• You can also think of acceleration as the second
derivative of position.
**Remember speed is the absolute value of velocity
E XAMP L E 1 : I F T H E P OSI T I ON OF
A PART I CL E I S G I V EN B Y T H E E QUATI ON ( W H E R E T I S
I N SE CONDS AND S I S I N FE E T ) :
s  f (t )  t  9t  15t
3
2
FI ND T H E FOL L OWI NG:
A. V E L OCI TY AT T I ME T
B . V E L OCIT Y AFT E R 3 SE CONDS
C . W H E N I S T H E PART I CL E AT R E ST ?
D. W H E N I S T H E PART I CL E MOV I NG I N A P OSI T I V E
DI R E CT ION?
E . W H E N I S I T MOV I NG I N A NE G ATI V E DI R E CT I ON?
F. FI ND T H E ACCE L ER ATION AT T I ME T.
G . FI ND ACCE L ERATI ON AFT E R 4 SE CONDS
H . W H E N I S T H E PART I CL E SP E E DING UP AND
SL OWI NG DOWN?
I . FI ND T H E TOTAL DI STANCE T R AV E LED B Y T H E
PART I CL E I N T H E FI R ST 6 SE CONDS
PART A: FIND THE VELOCITY AT TIME T
PART B: FIND THE VELOCITY AFTER 3
SECONDS.
Part A: Velocity is the first derivative of position.
f (t )  t 3  9t 2  15t
v(t )  f '(t )  3t  18t  15
2
Part B: Find the velocity after 3 seconds.
v(3)  3(3)  18(3)  15  12 ft / s
2
C . WHEN IS THE PARTICLE AT REST?
Part C: A particle is at rest when the velocity equals 0.
v(t )  3t 2  18t  15
0  3t 2  18t  15
0  3(t 2  6t  5)  3(t  5)(t  1)
t 5
t 1
The particle is at rest at 5 seconds and 1 second.
D. WHEN IS THE PARTICLE MOVING IN A
POSITIVE DIRECTION?
E. WHEN IS IT MOVING IN A NEGATIVE
DIRECTION?
PART D: THE PARTICLE IS MOVING IN A
POSITIVE DIRECTION WHEN THE
VELOCITY IS GREATER THAN 0.
v(t )  3t 2  18t  15
3t 2  18t  15  0
3(t  5)(t  1)  0
PART E: THE PARTICLE IS MOVING IN A
NEGATIVE DIRECTION WHEN THE
VELOCITY IS LESS THAN 0.
v(t )  3t 2  18t  15
3t  18t  15  0
2
t 1
3(t  5)(t  1)  0
t 5
1 t  5
F. FIND THE ACCELERATION AT TIME T.
G. FIND ACCELERATION AFTER 4 SECONDS
Part F: Acceleration is the derivative of velocity.
v(t )  3t  18t  15
2
a (t )  v '(t )  6t  18
Part G:
a (4)  6(4)  18  6 ft / s
2
H. WHEN IS THE PARTICLE SPEEDING UP
AND SLOWING DOWN?
Part H: The particle is speeding up when the velocity is positive and increasing (v
and a are both positive) and also when the velocity is negative and decreasing (v
and a are both negative). In other words, the particle speeds up when the
velocity and acceleration have the same sign.
Blue is the graph of the velocity function and red is the graph of the acceleration
function. We can look at the graph and make a sign table to find the intervals
when the particle is speeding up and slowing down.
Speeding Up:
Slowing Down:
I. FIND THE TOTAL DISTANCE TRAVELED BY
THE PARTICLE IN THE FIRST 6 SECONDS
Part I: Because the particle moves in both a positive and negative direction we
need to calculate the distances traveled during the time intervals [0,1], [1,5], and
[5,6] separately.
The distance traveled the first second is: f (1)  f (0)  7  0  7
The distance traveled from t=1 to t=5 is: f (5)  f (1)  25  7  32
The distance traveled from t=5 to t=6 is: f (6)  f (5)  18  25  7
7+32+7= 46 feet
Consider a graph of displacement (distance traveled) vs. time.
Average velocity can be found by
taking:
B
distance
(miles)
s
A
t
time (hours)
change in position s

change in time
t
Vave
s f  t  t   f  t 


t
t
The speedometer in your car does not measure average velocity, but instantaneous
velocity.
f  t  t   f  t 
ds
V t  
 lim
dt t 0
t
(The velocity at one moment in
time.)

Example:
Free Fall Equation
Gravitational
Constants:
1 2
h(t )   g t  v t  h
2
ft
g  32
sec 2
1
h(t )    32 t 2  v t  h
2
m
g  9.8
sec 2
h(t )  16 t 2  v t  h
ds
V (t ) 
 32 t  v0
dt
cm
g  980
sec 2
Speed is the absolute value of velocity.

Acceleration is the derivative of velocity.
2
d
h
dv
 2
a
dt
dt
If distance is in:
Velocity would be in:
Acceleration would be in:
v(t )  32t
example:
a (t )  32
feet
feet
sec
ft
ft
sec 
2
sec
sec

It is important to understand the relationship between a position graph, velocity and
acceleration:
acc neg
vel pos &
decreasing
acc neg
vel neg &
decreasing
acc zero
vel pos &
constant
distance
velocity
zero
acc pos
vel pos &
increasing
acc zero
vel neg &
constant
acc pos
vel neg &
increasing
acc zero,
velocity zero
time

Rates of Change:
Average rate of change =
f  x  h  f  x
h
Instantaneous rate of change =
f   x   lim
h 0
f  x  h  f  x
h
These definitions are true for any function.
( x does not have to represent time. )

HOMEWORK PAGE 231 #1-15 ODD