19. Rectilinear and Circular Motion

Adapted from: Classical single-variable calculus (position–velocity–acceleration on a line; free fall; angular velocity on a circle). Rewritten for teaching — not a verbatim textbook dump.

Motion is calculus with a physical story: derivatives of position give velocity; derivatives of velocity give acceleration.

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Rectilinear motion

Motion on a straight line. Place a coordinate $s$ on that line; the object's position at time $t$ is

$$s=f(t).$$

Fig. 19-1 — position s = f(t)

Between $t$ and $t+\Delta t$, the change in position is $f(t+\Delta t)-f(t)$. The average velocity is

$$\frac{f(t+\Delta t)-f(t)}{\Delta t}.$$

(This “distance” can be negative if the object moves toward decreasing $s$.)

The instantaneous velocity is the limit:

$$v=\lim_{\Delta t\to 0}\frac{f(t+\Delta t)-f(t)}{\Delta t}=f'(t)=\frac{ds}{dt}.$$

• If $v>0$ on an interval, $s$ is increasing (Theorem 13.7) — motion toward larger $s$.
• If $v<0$, motion toward smaller $s$.

Speed is $|v|$ — how fast, without direction (what a speedometer reports).

Acceleration is the rate of change of velocity:

$$a=\frac{dv}{dt}=\frac{d^{2}s}{dt^{2}}.$$

Example 19.1

$s=t^{2}-5t$ (miles, hours): $v=2t-5$ mi/h, $a=2$ mi/h². Velocity increases by $2$ mi/h every hour.

Turning around

A change of direction occurs at a relative extremum of $s$, which requires $v=ds/dt=0$. The converse fails: $s=t^{3}$ at $t=0$ has $v=0$ but no turn.

Example 19.2

$s=(t-2)^{2}$ (feet, seconds): $v=2(t-2)$, $a=2$.

• For $t<2$: $v<0$ — moving left.
• For $t>2$: $v>0$ — moving right.
• At $t=2$: $v=0$ — the object turns; it is not “standing still.” Standing still means $s$ is constant on a whole time interval.

Fig. 19-2 — s = (t−2)²

Fig. 19-3 — left then right

Check: velocity and speed

Quiz

Velocity vs speed

1 / 2

If v = ds/dt < 0, the object is…

Pick an answer to see if you are right.

Hands On: v and a from s(t)

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Motion under gravity (free fall)

Object thrown straight up or down (or released from rest); the only force is Earth's gravity. Put $s$ upward on the vertical line, with ground at $s=0$.

Physics: acceleration is constant

$$a\approx -32\ \mathrm{ft/s^{2}} \quad(\text{or }-9.8\ \mathrm{m/s^{2}}).$$

Negative because upward is positive — gravity decreases $v$.

Fig. 19-4 — upward +; a = −32 ft/s²

Since $dv/dt=-32$:

$$v=v_{0}-32t \tag{19.1}$$

with $v_{0}=v(0)$. Then $ds/dt=v$ gives

$$s=s_{0}+v_{0}t-16t^{2} \tag{19.2}$$

with $s_{0}=s(0)$.

(Reason: $D_t(v_{0}-32t)=-32=D_t v$, so the two velocities differ by a constant; they agree at $t=0$, so they are identical. Same argument for $s$.)

Check: free fall

Quiz

Free fall

1 / 2

With upward positive, a under gravity is…

Pick an answer to see if you are right.

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Circular motion

Particle $P$ on a circle: describe the motion by the central angle $\theta=f(t)$ in radians. If the radius is $r$,

$$x=r\cos\theta,\qquad y=r\sin\theta.$$

Angular velocity

$$\omega=\frac{d\theta}{dt}.$$

Angular acceleration

$$\alpha=\frac{d\omega}{dt}=\frac{d^{2}\theta}{dt^{2}}.$$

Check: circular motion

Quiz

Angular velocity

1 / 2

ω equals…

Pick an answer to see if you are right.

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Key takeaways

• Rectilinear: $v=ds/dt$, speed $=|v|$, $a=dv/dt=d^{2}s/dt^{2}$
• Sign of $v$ = direction; turns need $v=0$ at an extremum of $s$ (not every critical point)
• Free fall (upward +): $a=-32$ ft/s², $v=v_{0}-32t$, $s=s_{0}+v_{0}t-16t^{2}$
• Circular: $\omega=d\theta/dt$, $\alpha=d\omega/dt$; $(x,y)=(r\cos\theta,r\sin\theta)$

Further Reading

• 18. Inverse Trig Functions — prior chapter
• 18b. Inverse Trig — Practice — prior drills
• 13. Law of the Mean — $f'=0$ ⇒ constant (used to get (19.1)–(19.2))
• 14. Max and Min Values — extrema when $v=0$
• 19b. Motion — Practice — drills: distance, free fall, ω
• 20. Related Rates — next: linking dy/dt to dx/dt