Section 34a The difference quotient When we let h approach 0 we saw the rate at which a function w as changing at a particular point x Definition Instantaneous Rate of Change The ID: 437627
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Slide1
Velocity, Acceleration, Jerk
Section 3.4aSlide2
The difference quotient:
When we let
h
approach 0, we saw the rate at which a function
w
as changing at a particular point x…
Definition: Instantaneous Rate of Change
The
(instantaneous) rate of change
of f with respect to x at ais the derivative
provided the limit exists.Slide3
Example 1: Enlarging Circles
(a) Find the rate of change of the area
A
of a circle with respect
t
o its radius r.
Instantaneous rate of change of
A with respect to r :
(b) Evaluate the rate of change of A
at r = 5 and at r = 10.(c) If r is measured in inches and A is measured in square inches,
what units would be appropriate for
dA
/
dr
?
Rate at
r
= 5:
Rate at
r
= 10:
Units?
Square inches (of area)
per inch (of radius)
The rate of change gets bigger as
r
gets bigger!!!Slide4
Motion Along a Line
If an object is moving along an axis, we may know its position
s
on
that line as a function of time
t :
The
displacement
of the object over the interval
t to t + t :
The
average velocity
of the object over this time interval:
How would we find the object’s velocity at the exact instant
t
?Slide5
Definition: Instantaneous Velocity
The
(instantaneous) velocity
is the derivative of the position
function
s = f(
t) with respect to time. At time t
the velocity is
Definition: Speed
Speed is the absolute value of velocity.Speed =Slide6
Example 2: Reading a Velocity Graph
A particle moves along an axis, and its velocity is shown in the
graph below. When does the particle have maximum
speed
?
t (sec)
v (m/sec)
5
The particle moves forward for the first 10 seconds,
then moves backward for the next 8 seconds, stands
still for 4 seconds, and then moves forward again.
The particle achieves
its maximum speed at
about
t
= 15, while
moving backward.Slide7
Definition: Acceleration
Acceleration
is the derivative of velocity with respect to time.
If a body’s velocity at time
t
is v(
t) = ds/
dt, then the body’sacceleration at time t is
Definition: Jerk
Jerk is the derivative of acceleration with respect to time. If a
body’s position at time
t
is
s(t)
, the body’s jerk at time
t
isSlide8
Free-Fall Constants (Earth)
English units:
(
s
in feet)
Metric units:
(
s
in meters)Slide9
Example 3: Modeling Vertical Motion
A dynamite blast propels a heavy rock straight up with a launch
velocity of 160 ft/sec (about 109 mph). It reaches a height of
after
t
seconds.
Model the situation:
s
= 0
s
s
max
Height (ft)
v
= 0
In our model, velocity is positive on the
way up, and negative on the way down.
(a) How high does the rock go?
Find velocity at any time
t
:
ft/sec
The velocity is zero when:
secSlide10
Example 3: Modeling Vertical Motion
A dynamite blast propels a heavy rock straight up with a launch
velocity of 160 ft/sec (about 109 mph). It reaches a height of
after
t
seconds.
Model the situation:
s
= 0
s
s
max
Height (ft)
v
= 0
(a) How high does the rock go?
The maximum height of the rock is the
height at
t
= 5 sec:
ftSlide11
Example 3: Modeling Vertical Motion
A dynamite blast propels a heavy rock straight up with a launch
velocity of 160 ft/sec (about 109 mph). It reaches a height of
after
t
seconds.
s
= 0
s
s
max
Height (ft)
v
= 0
(b) What is the velocity and speed of the
rock when it is 256 ft above the ground
on the way up? On the way down?
256
t
= ?
ft/sec
ft/sec
At both instants, the speed of the rock is 96 ft/secSlide12
Example 3: Modeling Vertical Motion
A dynamite blast propels a heavy rock straight up with a launch
velocity of 160 ft/sec (about 109 mph). It reaches a height of
after
t
seconds.
s
= 0
s
s
max
Height (ft)
v
= 0
(c) What is the acceleration of the rock at
any time
t
during its flight?
256
t
= ?
The acceleration is always negative!!!Slide13
Example 3: Modeling Vertical Motion
A dynamite blast propels a heavy rock straight up with a launch
velocity of 160 ft/sec (about 109 mph). It reaches a height of
after
t
seconds.
s
= 0
s
s
max
Height (ft)
v
= 0
(d) When does the rock hit the ground?
256
t
= ?
The rock hits the ground
10 seconds after the blast.
Let’s graph the position, velocity, and
acceleration functions together in the
same window: [0,10] by [–160,400].Slide14
Example 4: A Moving Particle
A particle moves along a line so that its position at any time
is given by the function
where
s
is measured in meters and
t
is measured in seconds.
(a) Find the displacement during the first 5 seconds.
(b) Find the average velocity during the first 5 seconds.(c) Find the instantaneous velocity when t = 4.(d) Find the acceleration of the particle when
t
= 4.
(e) At what values of
t
does the particle change direction?
(f) Where is the particle when
s
is a minimum?Slide15
Example 4: A Moving Particle
A particle moves along a line so that its position at any time
is given by the function
where
s
is measured in meters and
t
is measured in seconds.
(a) Find the displacement during the first 5 seconds.
Displacement =
(b) Find the average velocity during the first 5 seconds.
Average Velocity =
(c) Find the instantaneous velocity when
t
= 4.Slide16
Example 4: A Moving Particle
A particle moves along a line so that its position at any time
is given by the function
where
s
is measured in meters and
t
is measured in seconds.
Acceleration =
(d) Find the acceleration of the particle when
t
= 4.
(e) At what values of
t
does the particle change direction?
(f) Where is the particle when
s
is a minimum?
Since acceleration is always positive,
the position
s
is at a minimum when
the particle changes direction, at
t
= 3/2.