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Applications of Integration: Position, Velocity and Acceleration
Related Topics
Wize University Calculus 1 Textbook > Applications of Differentiation for Science
Position, Velocity, and Acceleration
3 Activities
The velocity of a particle is given by
v
(
t
)
=
∫
0
5
f
(
x
)
d
x
v\left(t\right)=\int_0^5f\left(x\right)dx
v
(
t
)
=
∫
0
5
f
(
x
)
d
x
, the acceleration of this particle is
a
(
t
)
=
0
a\left(t\right)=0
a
(
t
)
=
0
a
(
t
)
=
f
(
x
)
a\left(t\right)=f\left(x\right)
a
(
t
)
=
f
(
x
)
a
(
t
)
=
f
(
5
)
a\left(t\right)=f\left(5\right)
a
(
t
)
=
f
(
5
)
a
(
t
)
=
f
(
5
)
−
f
(
0
)
a\left(t\right)=f\left(5\right)-f\left(0\right)
a
(
t
)
=
f
(
5
)
−
f
(
0
)
None of the above
I don't know
Check Submission
More Position, Velocity, and Acceleration Questions:
Applications of Integration: Position, Velocity and Acceleration
If the accelertion of a particle is given by
a
(
x
)
=
2
x
+
x
a\left(x\right)=2^x+x
a
(
x
)
=
2
x
+
x
, and
v
(
0
)
=
s
(
0
)
=
0
v\left(0\right)=s\left(0\right)=0
v
(
0
)
=
s
(
0
)
=
0
, then the displacement of the particle
s
(
x
)
s\left(x\right)
s
(
x
)
is
Practice: Velocity and Acceleration
A point
P
P
P
moves along the x-axis. Its position after
t
t
t
seconds is given by
x
=
2
t
3
−
21
t
2
+
60
t
x=2t^3-21t^2+60t
x
=
2
t
3
−
21
t
2
+
60
t
meters.
Practice: Velocity and Acceleration
Q.
\textbf{Q.}
Q.
A point
P
P
P
moves along the x-axis. Its position after
t
t
t
seconds is given by
x
=
2
t
3
−
21
t
2
+
60
t
x=2t^3-21t^2+60t
x
=
2
t
3
−
21
t
2
+
60
t
meters.
Practice: Velocity and Acceleration
Q.
\textbf{Q.}
Q.
A point
P
P
P
moves along the x-axis. Its position after
t
t
t
seconds is given by
x
=
2
t
3
−
21
t
2
+
60
t
x=2t^3-21t^2+60t
x
=
2
t
3
−
21
t
2
+
60
t
meters.
Applications of derivatives: Velocity and Direction
At time
t
≥
0
t\ge0
t
≥
0
, the velocity (in m/s) of a particle moving along a horizontal line is
v
(
t
)
=
t
2
−
2
t
+
2
v(t)=t^2-2t+2
v
(
t
)
=
t
2
−
2
t
+
2
Mean Value Theorem: Speed
In 2 hours, a car driver covered 150 km on a road with speed limit 65 km per hour. Did the car go over the speed limit? Answer using the mean-value theorem.
Applications of derivatives: Velocity and Direction
At time
t
≥
0
t\ge0
t
≥
0
, the velocity (in m/s) of a particle moving along a horizontal line is
v
(
t
)
=
t
2
−
2
t
+
2
v(t)=t^2-2t+2
v
(
t
)
=
t
2
−
2
t
+
2
Written Answer 3
The velocity of a protein inside a cell starting at the top and falling under the force of gravity is given by the differential equation
d
v
d
t
=
g
−
k
v
\displaystyle \frac{dv}{dt} = g - kv
d
t
d
v
=
g
−
k
v
where
g
≈
10
7
μ
m
/
s
2
g \approx 10^7 \mu m / s^2
g
≈
1
0
7
μ
m
/
s
2
is the gravitational acceleration.
k
v
kv
k
v
is the acceleration due to drag and
k
=
10
13
s
−
1
k = 10^{13} s^{-1}
k
=
1
0
13
s
−
1
. (Note that this model ignores diffusion of the protein which makes it a really bad model but we're exploring what gravity alone would do to a protein.
Applications of Integration: Position, Velocity and Acceleration
If the acceleartion of a particle is given by
a
(
x
)
=
2
x
+
x
a\left(x\right)=2^x+x
a
(
x
)
=
2
x
+
x
, and
v
(
0
)
=
s
(
0
)
=
0
v\left(0\right)=s\left(0\right)=0
v
(
0
)
=
s
(
0
)
=
0
, then the displacement of the particle
s
(
x
)
s\left(x\right)
s
(
x
)
is
Applications of Integration: Position, Velocity and Acceleration
If the displacement of a particle is given by
s
(
t
)
=
∫
0
t
x
2
sin
x
d
x
s\left(t\right)=\int_0^t\frac{x^2}{\sin x}dx
s
(
t
)
=
∫
0
t
s
i
n
x
x
2
d
x
, then the acceleration at at
t
=
π
2
t=\frac{\pi}{2}
t
=
2
π
is
Related Rates
An object is moving in such a way that its position is given by
x
(
t
)
=
2
t
3
−
t
x(t)=2t^3-t
x
(
t
)
=
2
t
3
−
t
Find the time
t
t
t
at which the object’s speed is equal to the average speed over the interval [0, 1].
Kinematics: Free fall
A ball at rest is dropped off a 150-m tall building. Ignoring air resistance, what will its speed be the instant it hits the ground?
Enter your answer as a positive value (speed) in m/s, do not include units.
Mean Value Theorem: Speed
In 2 hours, a car driver covered 150 km on a road with speed limit 65 km per hour. Did the car go over the speed limit? Answer using the mean-value theorem.
Practice: Velocity and Acceleration
Q.
\textbf{Q.}
Q.
A point
P
P
P
moves along the x-axis. Its position after
t
t
t
seconds is given by
x
=
2
t
3
−
21
t
2
+
60
t
x=2t^3-21t^2+60t
x
=
2
t
3
−
21
t
2
+
60
t
meters.
Applications of derivatives: Velocity and Direction
At time
t
≥
0
t\ge0
t
≥
0
, the velocity (in m/s) of a particle moving along a horizontal line is
v
(
t
)
=
t
2
−
2
t
+
2
v(t)=t^2-2t+2
v
(
t
)
=
t
2
−
2
t
+
2
An object is moving according to
x
(
t
)
=
t
4
−
8
t
2
+
2
x(t)=t^4-8t^2+2
x
(
t
)
=
t
4
−
8
t
2
+
2
. Find the time at which its acceleration is zero.
x
(
t
)
=
6
t
3
+
3
t
2
−
3
,
x(t)=6t^3+3t^2-3,
x
(
t
)
=
6
t
3
+
3
t
2
−
3
,
Find velocity and acceleration at t=2?
Applications of Integration
A particle moves in a straight line so that its velocity at time
t
t
t
is
v
(
t
)
=
t
+
1
v(t)=\sqrt{t+1}
v
(
t
)
=
t
+
1
. If its position at
t
=
8
t=8
t
=
8
is
s
(
8
)
=
20
s(8)=20
s
(
8
)
=
20
, find
s
(
9
)
s(9)
s
(
9
)
.