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Wize AP Calculus (BC) Textbook > Applications of Differentiation: Contextual

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.