AP Calculus AB
section 3.4
A dynamite blast propels a heavy
rock straight up with a launch
velocity of 160 ft/sec. The rock
reaches a height of s(t)=160t-16t2
feet after t seconds.
s(t)=160t-16t2
s ‘(t)=160-32t
s(t)
(feet)
time (seconds)
1. How far off the ground is the rock after 3
seconds?
1. How far off the ground is the rock after 3
seconds?
Find s(3), s(t)=160t-16t2
1. How far off the ground is the rock after 3
seconds?
Find s(3), s(t)=160t-16t2
s(3)=160(3)-16(32)
1. How far off the ground is the rock after 3
seconds?
Find s(3), s(t)=160t-16t2
s(3)=160(3)-16(32)
s(3)= 336 feet
2. How long does it take before the rock hits
the ground?
2. How long does it take before the rock hits
the ground?
The rock hits the ground when s(t)=0.
2. How long does it take before the rock hits
the ground?
The rock hits the ground when s(t)=0.
0=160t-16t2
2. How long does it take before the rock hits
the ground?
The rock hits the ground when s(t)=0
0=160t-16t2
0=16t(10-t)
2. How long does it take before the rock hits
the ground?
The rock hits the ground when s(t)=0
0=160t-16t2
0=16t(10-t)
0=16t
0=10-t
2. How long does it take before the rock hits
the ground?
The rock hits the ground when s(t)=0
0=160t-16t2
0=16t(10-t)
0=16t
0=10-t
t=0 seconds t=10 seconds
3. What is the rock’s maximum
height? When does this occur?
3. What is the rock’s maximum
height? When does this occur?
Maximum height occurs when
s’(t)=0.
3. What is the rock’s maximum
height? When does this occur?
Maximum height occurs when
s’(t)=0.
Let s’(t)=0
3. What is the rock’s maximum
height? When does this occur?
Maximum height occurs when
s’(t)=0.
Let s’(t)=0
0=160-32t
Recall: s(t)=160t-16t2
s’(t)=160-32t
3. What is the rock’s maximum
height? When does this occur?
Maximum height occurs when
s’(t)=0.
Let s’(t)=0
0=160-32t
32t=160
Recall: s(t)=160t-16t2
s’(t)=160-32t
3. What is the rock’s maximum
height? When does this occur?
Maximum height occurs when
s’(t)=0.
Let s’(t)=0
0=160-32t
32t=160
Recall:
s(t)=160t-16t2
s’(t)=160-32t
t=5 seconds
is when the max height occurs
3. What is the rock’s maximum
height? When does this occur?
Maximum height is s(5)
s(5)=160(5)-16(52)
Recall: s(t)=160t-16t2
s’(t)=160-32t
3. What is the rock’s maximum
height? When does this occur?
Maximum height is s(5)
s(5)=160(5)-16(52)
s(5)=400 feet
is the max height
Recall: s(t)=160t-16t2
s’(t)=160-32t
4. What is the rock’s average
velocity during the first 3 seconds?
4. What is the rock’s average
velocity during the first 3 seconds?
The average velocity on [0,3]:
f (3) − f (0)
A.V. =
3− 0
4. What is the rock’s average
velocity during the first 3 seconds?
The average velocity on [0,3]:
f (3) − f (0)
A.V. =
3− 0
336 − 0
A.V. =
3− 0
4. What is the rock’s average
velocity during the first 3 seconds?
The average velocity on [0,3]:
f (3) − f (0)
A.V. =
3− 0
336 − 0
A.V. =
3− 0
ft
A.V. = 112
sec
5. What is the rock’s velocity at
t=3?
5. What is the rock’s velocity at
t=3?
The rock’s velocity at t=3 refers to the
instantaneous velocity or s’(3).
5. What is the rock’s velocity at
t=3?
The rock’s velocity at t=3 refers to the
instantaneous velocity or s’(3).
s’(t)=160-32t
5. What is the rock’s velocity at
t=3?
The rock’s velocity at t=3 refers to the
instantaneous velocity or s’(3).
s’(t)=160-32t
s’(3)=160-32(3)
5. What is the rock’s velocity at
t=3?
The rock’s velocity at t=3 refers to the
instantaneous velocity or s’(3).
s’(t)=160-32t
s’(3)=160-32(3)
ft
s’(3)=64
sec
6. What is the rock’s acceleration
at t=3?
6. What is the rock’s acceleration
at t=3?
s(t) is the height of the rock as a function of time.
6. What is the rock’s acceleration
at t=3?
s(t) is the height of the rock as a function of time.
s’(t) is the velocity of the rock as a function of time.
6. What is the rock’s acceleration
at t=3?
s(t) is the height of the rock as a function of time.
s’(t) is the velocity of the rock as a function of time.
s’’(t) is the acceleration of the rock as a function
of time.
6. To find the acceleration at t=3,
we need to find s’’(t).
6. To find the acceleration at t=3,
we need to find s’’(t).
s(t) = 160t − 16t 2
6. To find the acceleration at t=3,
we need to find s’’(t).
s(t) = 160t − 16t 2
s '(t) = 160 − 32t
6. To find the acceleration at t=3,
we need to find s’’(t).
s(t) = 160t − 16t 2
s '(t) = 160 − 32t
s ''(t) = −32
6. To find the acceleration at t=3,
we need to find s’’(t).
s(t) = 160t − 16t 2
s '(t) = 160 − 32t
s ''(t) = −32
s’’(3)=-32
ft
2
sec
Does an acceleration of a constant
-32 ft/sec2 make sense?
Does an acceleration of a constant
-32 ft/sec2 make sense?
Velocity is the line with a
negative slope.
Does an acceleration of a constant
-32 ft/sec2 make sense?
Velocity is the line with a
negative slope.
7. What is the rock’s velocity at
t=6?
7. What is the rock’s velocity at
t=6?
s’(6)=160-32(6)
7. What is the rock’s velocity at
t=6?
s’(6)=160-32(6)
s’(6)=-32 feet per second
8. What is the rock’s speed at t=6?
8. What is the rock’s speed at t=6?
speed= |velocity|
8. What is the rock’s speed at t=6?
speed= |velocity|
If the velocity of the rock is -32 ft per second at
t=6, then………
8. What is the rock’s speed at t=6?
speed= |velocity|
If the velocity of the rock is -32 ft per second at
t=6, then………
the speed of the rock is 32 ft per second at t=6.
9. What is the acceleration of the
rock when the velocity is 0?
9. What is the acceleration of the
rock when the velocity is 0?
Let s’(t)=0 to find when the velocity is 0.
9. What is the acceleration of the
rock when the velocity is 0?
Let s’(t)=0 to find when the velocity is 0.
0=160-32t
9. What is the acceleration of the
rock when the velocity is 0?
Let s’(t)=0 to find when the velocity is 0.
0=160-32t
32t=160
9. What is the acceleration of the
rock when the velocity is 0?
Let s’(t)=0 to find when the velocity is 0.
0=160-32t
32t=160
t=5
9. What is the acceleration of the
rock when the velocity is 0?
Let s’(t)=0 to find when the velocity is 0.
0=160-32t
32t=160
t=5
And s’(5)=-32 feet per second per second
10. Is the rock speeding up or slowing down
at t=4, t=6?
10. Is the rock speeding up or slowing down
at t=4, t=6?
An object is speeding up if velocity and
acceleration have the same sign.
10. Is the rock speeding up or slowing down
at t=4, t=6?
An object is speeding up if velocity and
acceleration have the same sign.
Velocity=+ and Acceleration=+
10. Is the rock speeding up or slowing down
at t=4, t=6?
An object is speeding up if velocity and
acceleration have the same sign.
Velocity=+ and Acceleration=+
OR
10. Is the rock speeding up or slowing down
at t=4, t=6?
An object is speeding up if velocity and
acceleration have the same sign.
Velocity=+ and Acceleration=+
OR
Velocity= - and Acceleration= -
10. Is the rock speeding up or slowing down
at t=4, t=6?
An object is slowing down if velocity and
acceleration have the different signs.
10. Is the rock speeding up or slowing down
at t=4, t=6?
An object is slowing down if velocity and
acceleration have the different signs.
Velocity=+ and Acceleration=-
10. Is the rock speeding up or slowing down
at t=4, t=6?
An object is slowing down if velocity and
acceleration have the different signs.
Velocity=+ and Acceleration=OR
10. Is the rock speeding up or slowing down
at t=4, t=6?
An object is slowing down if velocity and
acceleration have the different signs.
Velocity=+ and Acceleration=OR
Velocity=- and Acceleration=+
10. Is the rock speeding up or slowing down
at t=4, t=6?
At t=4
10. Is the rock speeding up or slowing down
at t=4, t=6?
At t=4
s’(4)=160-32(4)=+
10. Is the rock speeding up or slowing down
at t=4, t=6?
At t=4
s’(4)=160-32(4)=+
s’’(4)=-32=-
10. Is the rock speeding up or slowing down
at t=4, t=6?
At t=4
s’(4)=160-32(4)=+
s’’(4)=-32=Since velocity and acceleration have different
signs the rock is slowing down at t=4.
10. Is the rock speeding up or
slowing down at t=4, t=6?
At t=6
10. Is the rock speeding up or
slowing down at t=4, t=6?
At t=6
s’(6)=160-32(6)=-
10. Is the rock speeding up or
slowing down at t=4, t=6?
At t=6
s’(6)=160-32(6)=s’’(6)=-32=-
10. Is the rock speeding up or
slowing down at t=4, t=6?
At t=6
s’(6)=160-32(6)=s’’(6)=-32=Since velocity and acceleration have the
same sign, the rock is speeding up at t=6.