4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration

4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration

When the changing quantity is the position of an object moving along a line, its derivatives have names of their own. Velocity is the rate of change of position and acceleration is the rate of change of velocity, so everything Topic 4.1 says about reading a derivative, with its sign, instant and units, applies directly to motion.

The enduring understanding for these topics is CHA-3: Derivatives allow us to solve real-world problems involving rates of change. The learning objective is CHA-3.B: Calculate rates of change in applied contexts. Its essential knowledge, CHA-3.B.1, states that the derivative can solve rectilinear-motion problems involving position, speed, velocity, and acceleration.

Straight-line motion: position, velocity, and acceleration

For a particle moving along a line, let $s(t)$ denote its position at time $t$. Position gives location relative to a chosen origin. Its first derivative is velocity:

$$ v(t)=s'(t), $$

and the derivative of velocity is acceleration:

$$ a(t)=v'(t)=s''(t). $$

Quantity	Meaning	Mathematical relationship	Typical units
Position $s(t)$	Location on the line	Given function	meters
Velocity $v(t)$	Signed rate of position change	$s'(t)$	meters per second
Speed	Magnitude of velocity	$\lvert v(t)\rvert$	meters per second
Acceleration $a(t)$	Rate of velocity change	$v'(t)=s''(t)$	meters per second squared

Velocity is signed, so it describes both how fast and which direction. A particle moves right when $v(t)>0$ and left when $v(t)<0$. Speed is never negative because it is $\lvert v(t)\rvert$.

Worked motion example

A particle moves along the $x$-axis with position

$$ x(t)=t^3-6t^2+9t, $$

where $x$ is measured in meters and $t$ in seconds. Differentiate:

$$ v(t)=x'(t)=3t^2-12t+9, $$

$$ a(t)=v'(t)=6t-12. $$

At $t=1$,

$$ v(1)=3-12+9=0, $$

so the particle is momentarily at rest. Its acceleration is

$$ a(1)=6(1)-12=-6\ \text{m/s}^2. $$

At $t=3$,

$$ v(3)=27-36+9=0, $$

so the particle is again at rest. To determine direction, factor velocity:

$$ v(t)=3(t-1)(t-3). $$

Thus $v(t)>0$ on $0<t<1$, $v(t)<0$ on $1<t<3$, and $v(t)>0$ for $t>3$. The particle moves right, then left, then right again.

Applying procedures with and without technology

The suggested skill for Topic 4.2 is 1.E: Apply appropriate mathematical rules or procedures, with and without technology. A graphing calculator may help graph $s(t)$, $v(t)$, and $a(t)$ or estimate a derivative numerically, but the interpretation must still include a sign, an instant, a quantity, and units.

Retrieval check: If $s(t)$ is measured in feet and $t$ in seconds, what do $s'(4)$ and $s''(4)$ measure? If $s'(4)=-12$, state its meaning in context.

Answer: $s'(4)$ is velocity in feet per second; $s''(4)$ is acceleration in feet per second squared. The value $s'(4)=-12$ means that at $t=4$ seconds, the particle is moving in the negative direction at $12$ feet per second.