Rates, motion, and units | Code Scholars Tutoring

Understand the idea

A derivative measures an instantaneous change in one quantity per unit of another. Velocity is the derivative of position, acceleration is the derivative of velocity, and speed is the magnitude of velocity. A particle speeds up where velocity and acceleration have the same nonzero sign. Interpret a second derivative using squared input units and describe average versus instantaneous rates accurately.

v(t)=s'(t),\quad a(t)=v'(t),\quad\text{speed}=|v(t)|

Worked example

For $\displaystyle s(t)=t^3-6t^2+9t$ , determine whether the particle speeds up at $\displaystyle t=4$ .

\displaystyle v(t)=3t^2-12t+9

and

\displaystyle a(t)=6t-12

. At

\displaystyle 4

\displaystyle v=9

and

\displaystyle a=12

. Both are positive, so speed is increasing. Positive acceleration alone would not suffice if velocity were negative.

Try it yourself

If $\displaystyle T'(3)=-2$ degrees per minute, interpret it.

Show solution and reasoning

At minute $\displaystyle 3$ , temperature is decreasing at an instantaneous rate of $\displaystyle 2$ degrees per minute. This does not say the temperature itself is negative.

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