High School
SAT
ACT
University
MCAT
LSAT
DAT

Written Answer 3

Related Topics

Wize University Calculus 1 Textbook > Applications of Differentiation for Science

Position, Velocity, and Acceleration

3 Activities

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.
What is the steady state velocity, vssv_{ss} , of the protein? Include units in your answer.
More Position, Velocity, and Acceleration Questions:
Applications of Integration: Position, Velocity and Acceleration
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
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
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).