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Wize University Calculus 1 Textbook > Applications of Differentiation for Science

Position, Velocity, and Acceleration

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Position, Velocity, and Acceleration


Since derivatives are rates of change, there is a natural relationship between position, velocity, and acceleration. Velocity is change in position while acceleration is change in velocity.

Position, Velocity, and Acceleration

Let
  • s(t)s\left(t\right)denote the position of a particle at time tt
  • v(t)v\left(t\right)denote the velocity of a particle at time tt
  • a(t)a\left(t\right)denote the acceleration of a particle at time tt
then

s′(t)=v(t)\boxed{s'\left(t\right)=v\left(t\right)}
and
s′′(t)=v′(t)=a(t)\boxed{s''\left(t\right)=v'\left(t\right)=a\left(t\right)}

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Example: Position, Velocity, and Acceleration

An object is moving and its position is governed by this equation x(t)=5t2−3t+1.x(t)=5t^2-3t+1.

What is the average velocity in time interval [0,2] seconds? Find the velocity at t=3t=3.


vavg=x(2)−x(0)2−0=(5∗4−6+1)−12=7 m/secv_{\text{avg}}=\frac{x(2)-x(0)}{2-0}=\frac{(5*4-6+1)-1}{2}=7\ \text{m/sec}

v(t)=x′(t)=10t−3→v(3)=10(3)−3=27 m/s\begin{array}{c} v(t)=x'(t)=10t-3\\ \text{} \\ \rightarrow v(3)=10(3)-3=27 \text{ m/s} \end{array}

Find velocity and acceleration at t=2t=2 for x(t)=6t3+3t2−3x(t)=6t^3+3t^2-3.



Extra Practice
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
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).