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Applications of Integration: Position, Velocity and Acceleration

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Wize University Calculus 1 Textbook > Applications of Differentiation for Science

Position, Velocity, and Acceleration

3 Activities

Wize University Calculus 1 Textbook > Integrals

Initial Conditions

3 Activities

If the accelertion of a particle is given by a(x)=2x+xa\left(x\right)=2^x+x, and v(0)=s(0)=0v\left(0\right)=s\left(0\right)=0, then the displacement of the particle s(x)s\left(x\right) is
More Position, Velocity, and Acceleration Questions:
Practice: Velocity and Acceleration
A point PP moves along the x-axis. Its position after tt seconds is given by x=2t3−21t2+60tx=2t^3-21t^2+60t meters.
Practice: Velocity and Acceleration
Q.\textbf{Q.} A point PP moves along the x-axis. Its position after tt seconds is given by x=2t3−21t2+60tx=2t^3-21t^2+60t meters.
Practice: Velocity and Acceleration
Q.\textbf{Q.} A point PP moves along the x-axis. Its position after tt seconds is given by x=2t3−21t2+60tx=2t^3-21t^2+60t meters.
Applications of derivatives: Velocity and Direction
At time t≥0t\ge0, the velocity (in m/s) of a particle moving along a horizontal line is v(t)=t2−2t+2v(t)=t^2-2t+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≥0t\ge0, the velocity (in m/s) of a particle moving along a horizontal line is v(t)=t2−2t+2v(t)=t^2-2t+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
dvdt=g−kv\displaystyle \frac{dv}{dt} = g - kv
where g≈107μm/s2g \approx 10^7 \mu m / s^2 is the gravitational acceleration. kvkv is the acceleration due to drag and k=1013s−1k = 10^{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)=2x+xa\left(x\right)=2^x+x, and v(0)=s(0)=0v\left(0\right)=s\left(0\right)=0, then the displacement of the particle s(x)s\left(x\right) is
Applications of Integration: Position, Velocity and Acceleration
If the displacement of a particle is given by s(t)=∫0tx2sin⁡xdxs\left(t\right)=\int_0^t\frac{x^2}{\sin x}dx, then the acceleration at at t=π2t=\frac{\pi}{2} is
Applications of Integration: Position, Velocity and Acceleration
The velocity of a particle is given by v(t)=∫05f(x)dxv\left(t\right)=\int_0^5f\left(x\right)dx, the acceleration of this particle is
Related Rates
An object is moving in such a way that its position is given by x(t)=2t3−tx(t)=2t^3-t Find the time ttat 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.} A point PP moves along the x-axis. Its position after tt seconds is given by x=2t3−21t2+60tx=2t^3-21t^2+60t meters.
Applications of derivatives: Velocity and Direction
At time t≥0t\ge0, the velocity (in m/s) of a particle moving along a horizontal line is v(t)=t2−2t+2v(t)=t^2-2t+2
An object is moving according to x(t)=t4−8t2+2x(t)=t^4-8t^2+2. Find the time at which its acceleration is zero.
x(t)=6t3+3t2−3,x(t)=6t^3+3t^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 tt is v(t)=t+1v(t)=\sqrt{t+1}. If its position at t=8t=8 is s(8)=20s(8)=20, find s(9)s(9).
More Initial Conditions Questions:
Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Practice: Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Practice: Indefinite Integral
Practice: Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Practice: Indefinite Integral (~F2018 Final Q26)
Practice: Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Exponential Functions: $f$ from $f''$
Given f′′(x)=5ex+2sin⁡xf''(x)=5e^x+2\sin x and given f(0)=0f(0)=0, f(π)=0f(\pi)=0, find f(x)f(x). Enter "f(x)=..."
Practice: Indefinite Integral
Find f(x)f\left(x\right), given that f′′(x)=x2−2+sin⁡xf''\left(x\right)=x^2-2+\sin x, f(0)=3f\left(0\right)=3 and f′(3)=1f'(3)=1
Practice: Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Applications of Integration: Position, Velocity and Acceleration
If the acceleartion of a particle is given by a(x)=2x+xa\left(x\right)=2^x+x, and v(0)=s(0)=0v\left(0\right)=s\left(0\right)=0, then the displacement of the particle s(x)s\left(x\right) is
Initial Conditions
Find the function y=f(x)y=f(x) if
y′′=3x−5/2y′(1)=1y(4)=6y''=3x^{-5/2} \quad y'(1)=1 \quad y(4)=6
Initial Conditions
Solve the initial value problem
y′=x2+1xy'=\frac{x^2+1}{\sqrt{x}}
y(1)=3y(1)=3
Initial Conditions
Solve the initial value problem
y′′=ex/2y′(0)=3y(0)=5y''=e^{x/2} \quad y'(0)=3 \quad y(0)=5
Initial Conditions
Find the function f(x)f(x) whose derivative is 4x−1/x24x-1/x^2 forx>0 x>0 and for which f(1)=−1f(1)=-1.
Solve the initial value problem
f(x)=4x+x22+5,F(1)=1/3\displaystyle f(x)=4x+\frac{x^2}{2}+5, F(1)=1/3
Solve the initial value problem
f(x)=x+x4x2,F(0)=2\displaystyle f(x)=\frac{\sqrt{x}+x^4}{x^2}, F(0)=2
Solve the initial value problem y′=sec⁡2x−e−x,    y(0)=2y'=\sec^2x-e^{-x},\ \ \ \ y\left(0\right)=2
Consider Newton's law of cooling dTdt=2−T10\displaystyle\frac{\text{d}T}{\text{d}t}=2-\frac{T}{10} . Determine T(t)T(t) if T(0)=100T(0)=100
Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Practice: Rearrange First
Q:\bf{Q:} Solve the initial value problem t2dydt+1=t3t^2\dfrac{dy}{dt} + 1 = t^3 with y(1)=1y(1) = 1
Q:\bf{Q:} Solve the initial value problem y′=x3+sin⁡2xy'=x^3+\sin2x given y(π2)=0y\left(\dfrac{\pi}{2}\right)=0
Q:\bf{Q:}Solve the initial value problem y′=sec⁡2x−e−xy'=\sec^2x-e^{-x} with y(0)=2y\left(0\right)=2
Solve the initial-value problem
y′=x2+1xy(1)=3\begin{aligned} y' &=\dfrac{x^2 + 1}{\sqrt{x}}\\ y(1) & = 3 \end{aligned}
Practice: From Second Derivative to Function
Q:\bf{Q:} Solve the initial-value problem y′′=sin⁡2xy'' = \sin 2x, y′(0)=3y'(0) = 3 , y(0)=5y(0) = 5.
Exponential Functions: $f$ from $f''$
Given f′′(x)=5ex+2sin⁡xf''(x)=5e^x+2\sin x and given f(0)=0f(0)=0, f(π)=0f(\pi)=0, find f(x)f(x). Enter "f(x)=..."
Solve the initial value problem
y′′=ex/2y′(0)=3y(0)=5\begin{aligned} y''& = e^{x/2}\\ y'(0)& = 3\\ y(0) &= 5 \end{aligned}
Solve the initial value problem y′=sec⁡2x−e−x,  y(0)=2y'=\sec^2x-e^{-x},\ \ y\left(0\right)=2.
Practice: Indefinite Integral
Practice: Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Find the antiderivative FF
f(x)=x+x4x2,F(1)=2\displaystyle f(x)=\frac{\sqrt{x}+x^4}{x^2}, F(1)=2
Find the antiderivative FF
f(x)=4x+x22+5,F(1)=1/3\displaystyle f(x)=4x+\frac{x^2}{2}+5, F(1)=1/3