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Math Problems
Calculus
Relate position, velocity, speed, and acceleration using derivatives
A bug is moving back and forth on a straight path. The velocity of the bug is given by
v
(
t
)
=
t
2
−
3
t
v(t)=t^{2}-3 t
v
(
t
)
=
t
2
−
3
t
. Find the average acceleration of the bug on the interval
[
1
,
4
]
[1,4]
[
1
,
4
]
.
\newline
∣
4
3
3
−
24
∣
∣
t
3
3
−
3
t
2
2
∣
1
4
\left|\frac{4^{3}}{3}-24\right|\left|\frac{t^{3}}{3}-\frac{3 t^{2}}{2}\right|_{1}^{4}
∣
∣
3
4
3
−
24
∣
∣
∣
∣
3
t
3
−
2
3
t
2
∣
∣
1
4
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Wed
17
17
17
Apr
\newline
Applications of integration: Unit test
\newline
A particle with velocity
v
(
t
)
=
3
t
v(t)=3 \sqrt{t}
v
(
t
)
=
3
t
, where
t
t
t
is time in seconds, moves in a straight line.
\newline
How far does the particle move from
t
=
1
t=1
t
=
1
to
t
=
4
t=4
t
=
4
seconds?
\newline
□
\square
□
units
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3
3
3
. A particle moves along a lines so that at time
t
t
t
where
0
≤
t
≤
π
0 \leq t \leq \pi
0
≤
t
≤
π
its position is given by
s
(
t
)
=
−
4
cos
t
−
t
2
2
+
10
s(t)=-4 \cos t-\frac{t^{2}}{2}+10
s
(
t
)
=
−
4
cos
t
−
2
t
2
+
10
. What is the velocity of the particle when its acceleration is zero?
\newline
cos
(
t
)
−
1
\cos (t)-1
cos
(
t
)
−
1
\newline
s
(
t
)
=
−
4
cos
t
−
t
2
2
+
10
v
(
t
)
=
4
sin
(
t
)
−
t
0
=
4
sin
(
t
)
−
t
t
=
2.4745768
a
(
2.47
)
=
4.14
u
/
u
\begin{array}{l} s(t)=-4 \cos t-\frac{t^{2}}{2}+10 \\ v(t)=4 \sin (t)-t \\ 0=4 \sin (t)-t \\ t=2.4745768 \\ a(2.47)=4.14 \mathrm{u} / \mathrm{u} \end{array}
s
(
t
)
=
−
4
cos
t
−
2
t
2
+
10
v
(
t
)
=
4
sin
(
t
)
−
t
0
=
4
sin
(
t
)
−
t
t
=
2.4745768
a
(
2.47
)
=
4.14
u
/
u
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QUESTION
6
6
6
\newline
With reference to the Figure: What is the rate of acceleration of the object between the
15
15
15
and
20
20
20
second marks (
15
15
15
-
20
20
20
seconds)? Note: Please express any deceleration answers by using a negative
(
−
)
(-)
(
−
)
symbol in front of your numerical answer. Note
1
1
1
: The units are not required to be included in this instance.
0
0
0
\qquad
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Basic Differentiation Rules and Rates of Change
\newline
121
121
121
\newline
05
05
05
. Velocity Verify that the average velocity over the time interval
[
t
0
−
Δ
t
,
t
0
+
Δ
t
]
\left[t_{0}-\Delta t, t_{0}+\Delta t\right]
[
t
0
−
Δ
t
,
t
0
+
Δ
t
]
is the same as the instantaneous velocity at
t
=
t
0
t=t_{0}
t
=
t
0
for the position function
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[In this question
i
i
i
and
j
j
j
are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin.] A ship
A
A
A
moves with constant velocity
(
3
i
−
10
j
)
kmh
−
1
(3i-10j)\,\text{kmh}^{-1}
(
3
i
−
10
j
)
kmh
−
1
. At time
t
t
t
hours, the position vector of
A
A
A
is
r
km
r\,\text{km}
r
km
. At time
t
=
0
t=0
t
=
0
,
A
A
A
is at the point with position vector
(
13
i
+
5
j
)
km
(13i+5j)\,\text{km}
(
13
i
+
5
j
)
km
. (a) Find
j
j
j
0
0
0
in terms of
t
t
t
.
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In this question
i
i
i
and
j
j
j
are horizontal unit vectors. A particle
P
P
P
of mass
2
kg
2\,\text{kg}
2
kg
moves under the action of two forces,
(
p
i
+
q
j
)
N
(p i + q j)\,\text{N}
(
p
i
+
q
j
)
N
and
(
2
q
i
+
p
j
)
N
(2q i + p j)\,\text{N}
(
2
q
i
+
p
j
)
N
, where
p
p
p
and
q
q
q
are constants. Given that the acceleration of
P
P
P
is
(
i
−
j
)
ms
−
2
(i - j)\,\text{ms}^{-2}
(
i
−
j
)
ms
−
2
(a) find the value of
p
p
p
and the value of
q
q
q
.
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Q
10
10
10
. A car has an initial velocity of
300
m
/
300 \mathrm{~m} /
300
m
/
minute. It travels a distance of
0.5
k
m
0.5 \mathrm{~km}
0.5
km
in
20
20
20
seconds. Use the appropriate formula to calculate the acceleration of the car in
m
/
s
2
\mathrm{m} / \mathrm{s}^{2}
m
/
s
2
.
\newline
v
=
u
+
a
t
s
=
u
t
+
1
2
a
t
2
v
2
=
u
2
+
2
a
s
v=u+a t \quad s=u t+\frac{1}{2} a t^{2} \quad v^{2}=u^{2}+2 a s
v
=
u
+
a
t
s
=
u
t
+
2
1
a
t
2
v
2
=
u
2
+
2
a
s
\newline
where
a
=
a=
a
=
constant acceleration,
u
=
u=
u
=
initial velocity,
v
=
v=
v
=
final velocity,
s
=
s=
s
=
displacement fror the position when
t
=
0
t=0
t
=
0
and
t
=
t=
t
=
time taken.
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According to Newton's Second Law of Motion, the sum of the forces that act on an object with a mass
m
m
m
that moves with an acceleration
a
a
a
is equal to
m
a
ma
ma
. An object whose mass is
80
80
80
grams has an acceleration of
20
20
20
meters per seconds squared. What calculation will give us the sum of the forces that act on the object, in Newtons (which are
kg
×
m/s
2
\text{kg} \times \text{m/s}^2
kg
×
m/s
2
)?
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Suppose you are driving a car along a straight road. Your velocity at any time
t
t
t
(in seconds) is given by the function
v
(
t
)
=
3
t
2
−
2
t
+
5
v(t)=3 t^{2}-2 t+5
v
(
t
)
=
3
t
2
−
2
t
+
5
meters per second. Find the total distance traveled by the car from
t
=
0
t=0
t
=
0
to
t
=
3
t=3
t
=
3
seconds.
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Force equals mass times acceleration.
\newline
F
=
m
a
\mathrm{F}=\mathrm{ma}
F
=
ma
\newline
Solve the equation for the mass,
m
m
m
.
\newline
m
=
[
□
]
[
□
]
\mathrm{m}=\frac{[\square]}{[\square]}
m
=
[
□
]
[
□
]
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A roller coaster is moving at
25
m
/
s
25 \mathrm{~m} / \mathrm{s}
25
m
/
s
at the bottom of a hill. Three seconds later it reaches the top of the hill moving at
10
m
/
s
10 \mathrm{~m} / \mathrm{s}
10
m
/
s
. What was the acceleration of the coaster?
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As a particle moves along the number line, its position at time
t
t
t
is
s
(
t
)
s(t)
s
(
t
)
, its velocity is
v
(
t
)
v(t)
v
(
t
)
, and its acceleration is
a
(
t
)
=
3
t
2
a(t)=3 t^{2}
a
(
t
)
=
3
t
2
.
\newline
If
v
(
0
)
=
3
v(0)=3
v
(
0
)
=
3
and
s
(
0
)
=
1
s(0)=1
s
(
0
)
=
1
, what is
s
(
2
)
?
s(2) ?
s
(
2
)?
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O. You drop an object off the roof of a
441
441
441
-foot building. The function
h
(
t
)
=
−
16
t
2
+
441
h(t)=-16 t 2+441
h
(
t
)
=
−
16
t
2
+
441
gives the objec leight
h
h
h
in feet above the ground after
t
t
t
seconds. After how many seconds does the object hit the grounc
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A train travelling at a uniform speed for
360
k
m
360 \mathrm{~km}
360
km
would have taken
48
48
48
minutes less to travel the same distance if its speed were
5
k
m
/
5 \mathrm{~km} /
5
km
/
hour more. Find the original speed of the train.
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11
11
11
. A body constrained to move along the
z
z
z
-axis of a coordinate system is subject to a constant force
F
F
F
given by
F
=
−
i
^
+
2
j
^
+
3
k
^
N
F=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} N
F
=
−
i
^
+
2
j
^
+
3
k
^
N
. Where
i
^
,
j
^
,
k
^
\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}
i
^
,
j
^
,
k
^
are unit vectors along the
x
,
y
x, y
x
,
y
and
z
z
z
axis of the system respectively. What is the work done by this force in moving the body a distance of
4
m
4 \mathrm{~m}
4
m
along the
z
\mathrm{z}
z
axis?
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A stunt woman falls into a net during the filming of an action movie. Assume she experiences an upward acceleration magnitude
a
a
a
, and a normal force magnitude
F
n
F_{n}
F
n
while touching the net.
\newline
Which of the following statements is correct for the stunt woman's mass
m
m
m
?
\newline
Consider upward as the positive direction.
\newline
Choose
1
1
1
answer:
\newline
(A)
m
=
a
+
g
F
n
m=\frac{a+g}{F_{n}}
m
=
F
n
a
+
g
\newline
(C)
m
=
−
F
n
a
+
g
m=\frac{-F_{n}}{a+g}
m
=
a
+
g
−
F
n
\newline
(C)
m
=
F
n
(
a
+
g
)
m=F_{n}(a+g)
m
=
F
n
(
a
+
g
)
\newline
(
1
1
1
)
m
=
F
n
a
+
g
m=\frac{F_{n}}{a+g}
m
=
a
+
g
F
n
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9
9
9
.
\newline
6
6
6
. The motion of a particle is described by the position function
s
(
t
)
=
2
t
3
−
15
t
2
+
33
t
+
17
,
t
>
0
s(t)=2 t^{3}-15 t^{2}+33 t+17, t>0
s
(
t
)
=
2
t
3
−
15
t
2
+
33
t
+
17
,
t
>
0
, where
t
t
t
is measured in seconds and
s
(
t
)
s(t)
s
(
t
)
in metres.
\newline
a) When is the particle at rest?
\newline
b) When is the acceleration of the particle positive?
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A pitcher throws a baseball towards home plate. The baseball, which has a mass of
145
g
145\,\text{g}
145
g
, approaches the plate with
122
J
122\,\text{J}
122
J
of kinetic energy. Calculate the speed of the baseball. Show your work.
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The radius of the base of a cone is increasing at a rate of
10
\mathbf{1 0}
10
meters per second.
\newline
The height of the cone is fixed at
6
6
6
meters.
\newline
At a certain instant, the radius is
1
1
1
meter.
\newline
What is the rate of change of the volume of the cone at that instant (in cubic meters per second)?
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The constant-pressure heat capacity of a sample of a perfect gas was found to vary with temperature according to the expression
C
P
/
(
J K
−
1
)
=
20.17
+
0.4001
(
T
/
K
)
C_P /(\text{J K}^{-1}) = 20.17 + 0.4001 (T/\text{K})
C
P
/
(
J K
−
1
)
=
20.17
+
0.4001
(
T
/
K
)
. Calculate
q
q
q
,
w
w
w
,
Δ
U
\Delta U
Δ
U
, and
Δ
H
\Delta H
Δ
H
when the temperature is raised from
0
∘
C
0 \, ^{\circ}\text{C}
0
∘
C
to
100
∘
C
100 \, ^{\circ}\text{C}
100
∘
C
(a) at constant pressure, and (b) at constant volume
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The speedometer readings of a car are shown below. Find the acceleration of the car and its displacement.
\newline
\newline
Time
\newline
Speedometer
\newline
\newline
9
9
9
:
25
25
25
am
\newline
36
km/h
36\,\text{km/h}
36
km/h
\newline
\newline
9
9
9
:
45
45
45
am
\newline
72
km/h
72\,\text{km/h}
72
km/h
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Define acceleration and give its SI unit. When is acceleration of a body negative? Give two examples of situations in which acceleration of the body is negative.
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Iry solving this:
\newline
A train accelerates to a speed of
20
m
/
s
20 \mathrm{~m} / \mathrm{s}
20
m
/
s
over a distance of
150
m
150 \mathrm{~m}
150
m
. Determine the acceleration (assume uniform) of the train.
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As a particle moves along the number line, its position at time
t
t
t
is
s
(
t
)
s(t)
s
(
t
)
, its velocity is
v
(
t
)
v(t)
v
(
t
)
, and its acceleration is
a
(
t
)
=
−
cos
t
a(t)=-\cos t
a
(
t
)
=
−
cos
t
.
\newline
If
v
(
π
)
=
2
v(\pi)=2
v
(
π
)
=
2
and
s
(
π
/
2
)
=
3
π
s(\pi / 2)=3 \pi
s
(
π
/2
)
=
3
π
, what is
s
(
0
)
s(0)
s
(
0
)
?
\newline
s
(
0
)
=
s(0)=
s
(
0
)
=
\newline
Stuck? Review related articles/videos or use a hint.
\newline
Report a problem
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(
5
5
5
pts) A triangle has a base that is decreasing at a rate of
11
c
m
/
s
11 \mathrm{~cm} / \mathrm{s}
11
cm
/
s
with the height being held constant. What is the rate of change of the area of the triangle if the height is
7
c
m
7 \mathrm{~cm}
7
cm
?
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the inclination (tilt ) of an amusement park ride is accelerating at a rate of
2160
degrees
/
minutes
2
2160 \, \text{degrees}/\text{minutes}^2
2160
degrees
/
minutes
2
what is the rides acceleration in
degrees
/
s
2
\text{degrees}/\text{s}^2
degrees
/
s
2
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k
=
1
2
m
v
2
k=\frac{1}{2} m v^{2}
k
=
2
1
m
v
2
\newline
Kinetic energy is defined to be the energy that an object possesses simply by being in motion. The formula used to calculate the kinetic energy,
k
k
k
, in joules
(
J
)
(\mathrm{J})
(
J
)
, of an object is shown, where
m
m
m
stands for the mass of the object in kilograms
(
k
g
)
(\mathrm{kg})
(
kg
)
, and
v
v
v
for its velocity in meters per second
(
m
s
)
\left(\frac{\mathrm{m}}{\mathrm{s}}\right)
(
s
m
)
. If a
650
k
g
650 \mathrm{~kg}
650
kg
object has a kinetic energy of approximately
100
,
000
J
100,000 \mathrm{~J}
100
,
000
J
, at what speed is the object moving?
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to increase the kinetic energy of an arrow by a factor of
6
6
6
, its speed should be
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the position of a particle moving in the xy plane is given by the parametric functions
x
(
t
)
=
e
(
n
t
)
x(t) = e^{(nt)}
x
(
t
)
=
e
(
n
t
)
and
y
(
t
)
=
3
(
t
−
1
)
y(t) = \frac{3}{(t-1)}
y
(
t
)
=
(
t
−
1
)
3
where
n
n
n
is a positive constant. If the line
x
+
2
y
=
6
x +2y = 6
x
+
2
y
=
6
is parallel to the line tangent to the path of the particle at
t
=
4
t = 4
t
=
4
, what is the value of
n
n
n
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Bill boards a Ferris wheel at the
3
3
3
-o'clock position and rides the Ferris wheel for several rotations. The Ferris wheel has a radius of
10
10
10
meters, and when Bill boards the Ferris wheel he is
16
16
16
meters above the ground. Imagine an angle with its vertex at the center of the Ferris wheel that subtends the path Bill travels.
\newline
a. Define a function
\newline
f
f
f
that expresses Bill's horizontal distance to the right of the center of the Ferris wheel (in meters) in terms of the number of radians
\newline
θ
\theta
θ
the angle has swept out since the ride started.
\newline
f
(
θ
)
=
f(\theta)=
f
(
θ
)
=
\newline
b. Define a function
\newline
g
g
g
that expresses Bill's distance above the ground (in meters) in terms of the number of radians
\newline
θ
\theta
θ
the angle has swept out since the ride started.
\newline
g
(
θ
)
=
g(\theta)=
g
(
θ
)
=
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Bill boards a Ferris wheel at the
3
3
3
-o'clock position and rides the Ferris wheel for several rotations. The Ferris wheel has a radius of
10
10
10
meters, and when Bill boards the Ferris wheel he is
16
16
16
meters above the ground. Imagine an angle with its vertex at the center of the Ferris wheel that subtends the path Bill travels.
\newline
a. Define a function
f
f
f
that expresses Bill's horizontal distance to the right of the center of the Ferris wheel (in meters) in terms of the number of radians
θ
\theta
θ
the angle has swept out since the ride started.
\newline
f
(
θ
)
=
f(\theta)=
f
(
θ
)
=
\newline
Preview
\newline
syntax error
\newline
b. Define a function
g
g
g
that expresses Bill's distance above the ground (in meters) in terms of the number of radians
θ
\theta
θ
the angle has swept out since the ride started.
\newline
g
(
θ
)
=
□
.
g(\theta)=\square \text {. }
g
(
θ
)
=
□
.
\newline
Preview
\newline
syntax error
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A particle moves along the
x
x
x
-axis with velocity given by
v
(
t
)
=
12
sin
(
4
t
)
v(t)=12 \sin (4 t)
v
(
t
)
=
12
sin
(
4
t
)
for time
t
≥
0
t \geq 0
t
≥
0
. If the particle is at position
x
=
1
x=1
x
=
1
at time
t
=
0
t=0
t
=
0
, what is the position of the particle at time
t
=
π
8
t=\frac{\pi}{8}
t
=
8
π
?
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The period
T
T
T
(in seconds) of a pendulum is given by
T
=
2
π
(
L
32
)
T=2\pi\sqrt{(\frac{L}{32})}
T
=
2
π
(
32
L
)
, where
L
L
L
stands for the length (in feet) of the pendulum. If
π
=
3.14
\pi=3.14
π
=
3.14
, and the period is
9.42
9.42
9.42
seconds, what is the length? The length of the pendulum is in feet.
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Computation.
\newline
A crate of mass
248
kg
248\,\text{kg}
248
kg
is on a rough ramp, inclined at
3
0
∘
30^{\circ}
3
0
∘
above the horizontal. The coefficient of kinetic friction between the crate and ramp is
0.237
0.237
0.237
. A horizontal force of magnitude
F
=
5
,
214
N
F=5,214\,\text{N}
F
=
5
,
214
N
is applied to the crate as in the figure, pushing it up the ramp. What is the crate's acceleration, taking "up the ramp" to be the positive direction.
\newline
https://erau.instructure.com/courses/
163972
163972
163972
/quizzes/
568764
568764
568764
/history?version=
2
2
2
\newline
3
7
\frac{3}{7}
7
3
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A particle moves along a straight line. Its speed is inversely proportional to the square of the distance,
S
S
S
, it has traveled. Which equation describes this relationship?
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Acceleration
\newline
Quiz - Acceleration
\newline
A roller coaster is moving at
25
m
s
25\frac{m}{s}
25
s
m
at the bottom of a hill. Three seconds later it reaches the top of the hill moving at
10
m
s
10\frac{m}{s}
10
s
m
. What was the acceleration of the coaster?
\newline
A.
−
11.7
m
s
2
-11.7\frac{m}{s^{2}}
−
11.7
s
2
m
\newline
B.
−
5
m
s
2
-5\frac{m}{s^{2}}
−
5
s
2
m
\newline
C.
−
11.7
m
s
2
-11.7\frac{m}{s^{2}}
−
11.7
s
2
m
\newline
D.
5
m
s
2
5\frac{m}{s^{2}}
5
s
2
m
\newline
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Neil Armstrong dropped a hammer on the moon. If it took
4
s
4\,\text{s}
4
s
to fall and the final speed was
8
m/s
8\,\text{m/s}
8
m/s
, what is the acceleration due to gravity on the moon?
\newline
A.
9.8
m/s
2
9.8\,\text{m/s}^2
9.8
m/s
2
\newline
B.
32
m/s
2
32\,\text{m/s}^2
32
m/s
2
\newline
C.
4
m/s
2
4\,\text{m/s}^2
4
m/s
2
\newline
D.
2
m/s
2
2\,\text{m/s}^2
2
m/s
2
\newline
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A cart rolling down an incline for
5.0
5.0
5.0
seconds has an acceleration of
4.0
m/s
2
4.0\,\text{m/s}^{2}
4.0
m/s
2
. If the cart has a beginning speed of
2.0
m/s
2.0\,\text{m/s}
2.0
m/s
, what is its final speed?
\newline
A.
22
m/s
22\,\text{m/s}
22
m/s
\newline
B.
20
m/s
20\,\text{m/s}
20
m/s
\newline
C.
18
m/s
18\,\text{m/s}
18
m/s
\newline
D.
24
m/s
24\,\text{m/s}
24
m/s
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A roller coasters velocity at the top of the hill is
10
m
s
10\frac{m}{s}
10
s
m
. Two seconds later it reaches the bottom of the hill with a velocity of
26
m
s
26\frac{m}{s}
26
s
m
. What is the acceleration of the coaster?
\newline
A.
18
m
s
2
18\frac{m}{s^{2}}
18
s
2
m
\newline
B.
10
m
s
2
10\frac{m}{s^{2}}
10
s
2
m
\newline
C.
8
m
s
2
8\frac{m}{s^{2}}
8
s
2
m
\newline
D.
32
m
s
2
32\frac{m}{s^{2}}
32
s
2
m
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A particle moves along the
x
x
x
-axis so that at time
t
≥
0
t \geq 0
t
≥
0
its velocity is given by
v
(
t
)
=
6
t
2
−
24
t
+
18
v(t)=6 t^{2}-24 t+18
v
(
t
)
=
6
t
2
−
24
t
+
18
. Determine all intervals when the particle is moving to the right.
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A particle moves along the
x
x
x
-axis so that at time
t
≥
0
t \geq 0
t
≥
0
its position is given by
x
(
t
)
=
t
5
−
5
t
4
x(t)=t^{5}-5 t^{4}
x
(
t
)
=
t
5
−
5
t
4
. Determine all intervals when the particle is moving to the right.
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A particle moves along the
x
x
x
-axis so that at time
t
≥
0
t \geq 0
t
≥
0
its position is given by
x
(
t
)
=
t
5
−
5
t
4
x(t)=t^{5}-5 t^{4}
x
(
t
)
=
t
5
−
5
t
4
. Determine all intervals when the particle is moving to the right.
\newline
Moving right
→
v
(
t
)
>
0
\text { Moving right } \rightarrow v(t)>0
Moving right
→
v
(
t
)
>
0
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A particle moves along the
x
x
x
-axis so that at time
t
≥
0
t \geq 0
t
≥
0
its position is given by
x
(
t
)
=
−
5
t
4
+
30
t
2
x(t)=-5 t^{4}+30 t^{2}
x
(
t
)
=
−
5
t
4
+
30
t
2
. Determine all intervals when the speed of the particle is increasing.
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Example Problems
\newline
A particle moves along the
x
x
x
-axis so that at time
t
≥
0
t \geq 0
t
≥
0
its position is given by
x
(
t
)
=
−
5
t
4
+
30
t
2
x(t)=-5 t^{4}+30 t^{2}
x
(
t
)
=
−
5
t
4
+
30
t
2
. Determine all intervals when the speed of the particle is increasing.
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Word Poblem
\newline
A man flies a kite with a
100
100
100
foot string. The angle of elevation of the string is
5
2
∘
52^{\circ}
5
2
∘
.
\newline
What is the height of the kite in the sky?
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Word poslem
\newline
1
1
1
. A man flies a kite with a
100
100
100
foot string. The angle of elevation of the string is
5
2
∘
52^{\circ}
5
2
∘
.
\newline
What is the height of the kite in the sky?
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MODELING REAL LIFE The function
f
(
x
)
=
−
0.003
(
x
−
200
)
2
+
123
f(x)=-0.003(x-200)^{2}+123
f
(
x
)
=
−
0.003
(
x
−
200
)
2
+
123
represents the path of the baseball after being hit, where
x
x
x
is the horizontal distance (in feet) and
f
(
x
)
f(x)
f
(
x
)
is the height (in feet).
\newline
a. Identify the graph of the function.
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A particle moves along the
x
x
x
-axis so that at time
t
≥
0
t \geq 0
t
≥
0
its velocity is given by
v
(
t
)
=
18
t
2
−
72
t
+
54
v(t)=18 t^{2}-72 t+54
v
(
t
)
=
18
t
2
−
72
t
+
54
. Determine all intervals when the speed of the particle is increasing.
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A particle moves along the
x
x
x
-axis so that at time
t
≥
0
t \geq 0
t
≥
0
its velocity is given by
v
(
t
)
=
6
t
2
−
48
t
+
90
v(t)=6 t^{2}-48 t+90
v
(
t
)
=
6
t
2
−
48
t
+
90
. Determine all intervals when the speed of the particle is decreasing.
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