3.4 Velocity, Speed, and Rates of Change
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.)

Velocity is the first derivative of position.

Example:
Free Fall Equation
Gravitational
Constants:
1 2
s g t
2
ft
g  32
sec 2
1
s   32 t 2
2
m
g  9.8
sec 2
s  16 t 2
ds
V
 32 t
dt
cm
g  980
sec 2
Speed is the absolute value of velocity.

A fly ball is hit vertically upward (a pop-fly). It’s position is
given by the equation s(t) = -5t2+30t, where the origin is at
ground level and the positive direction is up.
a) Find the max height of the baseball.
Grade 10: Complete the
Square
Calculus: Use derivatives
Find where v(t)= s’(t) = 0
s(t)=
– 6t)
v(t)= -10t + 30
Set v(t) = 0 and solve
= -5(t2 – 6t +9 – 9)
-10t + 30 = 0
2
= -5[(t – 3) – 9]
t=3
Sub t = 3 into s(t) to find height:
= -5(t – 3)2 +45
s(3) = -5(3)2 + 30(3)
= 45
 The max height of the ball is 45m
-5(t2
b) When will the ball hit the ground?
Set s(t) = 0 and solve for t
-5t2+30t = 0
-5t(t - 6) = 0
t = 0 or t = 6

It took 6 seconds to return to the ground.

c) What is the velocity when it hits the ground?
v(6) = -10(6) + 30
= -30
It was travelling 30m/s when it hit the ground.

The Cheetah Problem
The position function of a cheetah moving across level ground in a straight line chasing
after prey is given by the equation s(t) = t3-15t2+63t, where t is in seconds and
s(t) is in metres.
a)
What is the cheetah’s velocity at 1s?4s?8s?
b)
When is it momentarily stopped? Did it turn around?
c)
When is its motion positive? negative?
d)
Find the position of the cheetah after 10s?
e)
Find the total distance travelled after 10s.
The position function of an object moving in a straight line is
s(t) = 3t2 – 0.25t4. At time t = 3, is the object moving towards
the origin or away from the origin?
Solution:
The velocity of the object is: v(t) = s’(t)
= 6t – t3
At time t = 3,
27
s(3) =
and v(3) = - 9
4
In the diagram, let the positive direction be to the right.
Thus, s(3) >0 implies the object is to the right of the origin
and v(3) < 0 implies the object is moving to the left
(in this case, to the origin) when t = 3.
27
4
0
Direction of motion at t = 3
s
Acceleration is the derivative of velocity.
2
d
s
dv
 2
a
dt
dt
example:
v  32t
a  32
If distance is in:
feet
Velocity would be in:
feet
sec
ft
Acceleration would be in: sec  ft
2
sec
sec
metre
metre
sec
metre
sec  metre
sec
sec 2

Starting at time t = 0, a dragster accelerates down a strip
and then brakes and comes to a stop. Its position function
s(t) is s(t) = 6t2 – 0.2t3.
a)
b)
c)
After how many seconds does the dragster stop?
What distance does the dragster travel?
At what time does the braking commence?
Solutions:
a)
v(t) = 12t – 0.6t2. To stop v(t) = 0
12t – 0.6t2=0 so that t = 0 or t = 20. It stops after 20s.
b)
It travels from s(0) = 0 to s(20)= 800. Since it did not reverse
direction, the distance travelled was 800m.
c)
a(t) = v’(t) = 12 – 1.2t = 1.2(10 – t)
Note: when t<10 then a(t) >0 and that if t>10 then a(t) <0. Thus the
dragster accelerates from 0s to 10s and brakes from 10s to 20s.
The braking begins after 10s.

A ball thrown in the air slows as it rises and speeds up as it falls
Negative acceleration means that the ball slows down as it rises.
It indicates that the velocity is decreasing.
dv
a 
0
dt
dv
0
Positive acceleration a 
dt
is increasing.
means that velocity
An object is accelerating if a(t) x v(t) > 0
decelerating if a(t) x v(t) < 0
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
f  x  h  f  x
Instantaneous rate of change = f   x   lim
h 0
h
These definitions are true for any function.
( x does not have to represent time. )

Example 1:
For a circle:
A   r2
dA d
  r2
dr dr
dA
 2 r
dr
Instantaneous rate of change of the area with
respect to the radius.
dA  2 r dr
For tree ring growth, if the change in area is constant then dr
must get smaller as r gets larger.

from Economics:
The instantaneous rate of change of cost with respect to the
number of items produced is called the marginal cost.
Marginal cost is the first derivative of the cost function, and
represents an approximation of the cost of producing one
more unit.
Marginal revenue is the first derivative of the revenue function,
sometimes given as R(x) = xp(x) where p(x) is the price per
unit (p(x) is sometimes called the price function or demand
function.) and x is the number of items.

Example:
Suppose you are running a company that
assembles stoves. You have been keeping
Track of your expenses, but it is a small company
and you only have partial figures.
# Stoves 5
8
10
12
15
Cost
1588
1978
2368
2998
1000
If you are currently producing 10 stoves, how
much will the 11th stove cost?
If we use the slope of the secant around 10 we can use it
to estimate the cost of making the 11th stove.
2368 − 1588
Slope=
=$195
12 − 4
marginal cost

Marginal cost is a linear approximation of a curved
function. For large values it gives a good approximation
of the cost of producing the next item.
This approximation is based on the idea of
local linearization. It is the process of using a tangent or a
secant to approximate a value of a function in close
proximity to a known point.
195
1
10
marginal cost
11

Example:
c  x   x3  6 x 2  15x
Suppose it costs:
to produce x stoves.
c  x   3x2 12x  15
If you are currently producing 10 stoves, the
11th stove will have a marginal cost of
approximately:
2
Note that this is not a great
approximation – Don’t let that
bother you.
c 10  3 10 12 10  15
 300 120 15
 $195
The actual cost is: C 11  C 10
marginal cost
 113  6 112  15 11  103  6 102  15 10 
 770  550  $220
actual cost

Guidelines for Successful Problem
Solving with Related Rates
1. Make a sketch and label known quantities
2. Introduce variables to rep. quantities that
change (let x rep…)
3. Identify what you are to find
4. Find an equation (area, volume, pythag…)
5. Find the derivative
6. Substitute and solve
7. Conclude in the context of the problem
