19b. Rectilinear and Circular Motion — Practice

Drill sheet for 19. Rectilinear & Circular Motion. Book-style solved ideas as guided checks and regenerable worksheets.

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1. Evaluate $v$ and $a$ at a time

Book. $s=\tfrac12 t^{3}-2t$: at $t=2$, $v=4$, $a=6$.

v and a at a given t

Differentiate s, then plug in

(1)For $s=\tfrac12 t^{3}-2t$, find $v$ and $a$ at $t=2$.

$v=\tfrac32 t^{2}-2$ ⇒ $v(2)=4$.

$a=3t$ ⇒ $a(2)=6$.

$v=4$; $a=6$.

(2)For $s=t^{3}-1t$, find $v$ and $a$ at $t=2$.

$v=3t^{2}-1$ ⇒ $v(2)=12-1$.

$a=6t$ ⇒ $a(2)=12$.

$v=11$; $a=12$.

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2. Full analysis of a cubic path

Book. $s=t^{3}-6t^{2}+9t+4$:

$$v=3(t-1)(t-3),\qquad a=6(t-2).$$

Question	Answer
$v=0$	$t=1$: $s=8$, $a=-6$; $t=3$: $s=4$, $a=6$
$a=0$	$t=2$: $s=6$, $v=-3$
$s$ increasing	$t<1$ or $t>3$
$v$ increasing	$t>2$
Direction changes	$t=1$ and $t=3$

Intervals from v and a

Factor, then read sign charts

(1)For $s=t^{3}-6t^{2}+9t+4$: when is $s$ increasing? When does direction change?

$v=3(t-1)(t-3)$; $s$ increasing for $t<1$ and $t>3$.

Turns at $t=1$ and $t=3$ ($v=0$, $a\neq 0$).

Inc. on $(-\infty,1)\cup(3,\infty)$; turns at $1,3$.

(2)For $s=t^{3}-9t^{2}+24t$: when is $s$ decreasing? When is $v$ increasing?

$v=3(t-2)(t-4)$ ⇒ $s$ decreasing on $(2,4)$.

$a=6(t-3)$ ⇒ $v$ increasing for $t>3$.

Dec. on $(2,4)$; $v$ inc. for $t>3$.

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3. Total distance with turns

Book. $s=t^{3}-9t^{2}+24t$: turns at $t=2,4$.

$s(0)=0\to s(2)=20\to s(4)=16\to s(5)=20$: total distance $20+4+4=28$ ft.

Fig. 19-5 — total distance 28 ft

Total distance traveled

Sum |Δs| between consecutive critical times

(1)Particle: $s(0)=3$, moves left to $s(2.5)=-27/16$, then to $s(3)=0$. Total distance on $[0,3]$?

$3-(-27/16)+|0-(-27/16)|=3+27/16+27/16=51/8$.

$51/8$.

(2)Particle: $s(0)=3$, moves left to $s(2.5)=-27/16$, then to $s(3)=0$. Total distance on $[0,3]$?

$3-(-27/16)+|0-(-27/16)|=3+27/16+27/16=51/8$.

$51/8$.

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4. Speed vs velocity

Book. $v=2(t-1)^{2}(2t-5)$, $a=12(t-1)(t-2)$.

Speed $|v|$ increases when $v$ and $a$ have the same sign; decreases when they have opposite signs. Direction changes at $t=2.5$ (simple zero of $v$ that is an extremum of $s$). Total distance on $[0,3]$ is $51/8$.

Fig. 19-6 — total distance 51/8

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5. Projected upward

Book. $s=112t-16t^{2}$: $v=112-32t$, $a=-32$.

• $t=3$: rising at $16$ ft/s; $t=4$: falling at $16$ ft/s
• Peak: $v=0$ ⇒ $t=3.5$, $s=196$ ft
• Height $96$ ft at $t=1$ (rising) and $t=6$ (falling)

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6. Circular motion numbers

Book. $\theta=t^{3}/50-t$: at $t=10$, $\theta=10$, $\omega=5$, $\alpha=6/5$.

θ, ω, α

Differentiate the angle function

(1)If $\theta=2t^{2}$, find $\omega$ and $\alpha$ at $t=2$.

$\omega=2\cdot2t=4t$ ⇒ $\omega(2)=8$.

$\alpha=4$ (constant).

$\omega=8$; $\alpha=4$.

(2)If $\theta=2t^{2}$, find $\omega$ and $\alpha$ at $t=2$.

$\omega=2\cdot2t=4t$ ⇒ $\omega(2)=8$.

$\alpha=4$ (constant).

$\omega=8$; $\alpha=4$.

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7–8. Dropped stone and rocket

Book 7. Dropped from $1024$ ft: $s=1024-16t^{2}$ ⇒ hit at $t=8$ with speed $256$ ft/s $=174\tfrac{6}{11}$ mi/h via

$$x\ \mathrm{ft/s}=\frac{15}{22}x\ \mathrm{mi/h}. \tag{19.3}$$

Book 8. Rocket $v_0=192$: peak at $t=6$, height $576$ ft; back at $t=12$ with speed $192$ ft/s (symmetric flight).

Free-fall applications

Peak, impact, unit conversion

(1)Dropped from $1024$ ft ($v_0=0$). When does it hit, and with what speed?

$1024-16t^{2}=0$ ⇒ $t=8$.

$v=-32\cdot 8=-256$ ft/s; speed $256$ ft/s.

$t=8$ s; speed $256$ ft/s.

(2)Dropped from $1024$ ft ($v_0=0$). When does it hit, and with what speed?

$1024-16t^{2}=0$ ⇒ $t=8$.

$v=-32\cdot 8=-256$ ft/s; speed $256$ ft/s.

$t=8$ s; speed $256$ ft/s.

(3)Rocket: $v_0=192$ ft/s from ground. Max height time and height? Return time?

$v=192-32t=0$ ⇒ $t=6$, $s=576$ ft.

Ground again at $t=12$; impact speed $192$ ft/s.

$6$ s; $576$ ft; $12$ s.

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Supplementary Problems

9. Prove the speed criterion

Speed $S=|v|$. If $v>0$ then $S=v$ and $dS/dt=a$. If $v<0$ then $S=-v$ and $dS/dt=-a$. So $S$ increases when $v$ and $a$ have the same sign, and decreases when they have opposite signs.

10. Snapshots along $s=t^{3}-6t^{2}+9t$

At each $t$, report $s$, direction (sign of $v$), and whether speed is increasing.

$t$	$s$	motion	speed
$1/2$	$25/8$	right	decreasing
$3/2$	$27/8$	left	increasing
$5/2$	$5/8$	left	decreasing
$4$	$4$	right	increasing

Position · direction · speed trend

Problem 10 style

(1)For $s=t^{3}-6t^{2}+9t$: at $t=1/2$ and $t=3/2$, give direction and whether speed is increasing.

$v=3(t-1)(t-3)$, $a=6(t-2)$.

$t=1/2$: $v>0$, $a<0$ ⇒ right, speed decreasing.

$t=3/2$: $v<0$, $a<0$ ⇒ left, speed increasing.

Right, slowing; left, speeding up.

(2)For $s=t^{3}-6t^{2}+9t$ at $t=4$: $s$, $v$, and speed trend?

$s=4$; $v=9$ (right); $a>0$ ⇒ speed increasing.

$s=4$; right at $9$ ft/s; speeding up.

11–12. Reverse gear / direction changes

Book 11. Locomotive $s=3t^{4}-44t^{3}+144t^{2}$ is in reverse on $3<t<8$.

Book 12. Direction changes: (a) $t=2,4$; (b) none; (c) $t=1,3$; (d) $t=1$.

When does direction change?

Factor v; discard double roots that do not reverse

(1)Direction changes for $s=2t^{3}-12t^{2}+18t-5$?

$v=6(t-1)(t-3)$; turns at $1$ and $3$.

$t=1$ and $t=3$.

(2)Direction changes for $s=2t^{3}-12t^{2}+18t-5$?

$v=6(t-1)(t-3)$; turns at $1$ and $3$.

$t=1$ and $t=3$.

13–14 / 16–17 / 20. More free fall

Book 13. $s=64t-16t^{2}$: at $48$ ft of rise, $v=v_0/2$.

Book 14. From roof, $s=96t-16t^{2}$; street $112$ ft below. At $t=2$: $240$ ft above street, $32$ ft/s up; impact $v=-128$ ft/s.

Book 16–17. Drop $144$ ft → $3$ s; drop $100$ ft → $54\tfrac{6}{11}$ mi/h.

Book 20. Ground-to-ground in $15$ s ⇒ $v_0=240$ ft/s.

Free-fall supplementary

Half-velocity, roofs, wells, unit conversion

(1)Ball from roof: $s=96t-16t^{2}$ relative to roof; street is $112$ ft below. Velocity when it hits the street?

Street: $s=-112=96t-16t^{2}$ ⇒ $16t^{2}-96t-112=0$ ⇒ $t^{2}-6t-7=0$ ⇒ $t=7$ ($t\ge 0$).

$v=96-32t$; $v(7)=-128$ ft/s.

$-128$ ft/s (book uses $s=96t-16t^{2}$ from roof).

(2)Dropped from a $10$-story building ($100$ ft). Impact speed in mi/h?

$100-16t^{2}=0$ ⇒ $t=2.5$; $|v|=80$ ft/s.

$\tfrac{15}{22}\cdot 80=54\tfrac{6}{11}$ mi/h.

$54\tfrac{6}{11}$ mi/h.

(3)For $s=64t-16t^{2}$, show velocity is halved after the first $48$ ft of rise.

$v=64-32t$; $v_0=64$.

$48=64t-16t^{2}$ ⇒ $t^{2}-4t+3=0$ ⇒ $t=1$ (rising).

$v(1)=32=v_0/2$.

At $s=48$, $v=32=v_0/2$.

15 / 18–19. Wheel and road distance

Book 15. $\theta=128t-12t^{2}$ at $t=3$: $\omega=56$, $\alpha=-24$.

Book 18. $s=8t^{3}-12t^{2}+6t-1$ on $[0,1]$: distance $2$ mi ($v\ge 0$).

Book 19. $s=5t-t^{2}$ on $[0,3]$: turn at $t=2.5$; distance $6.5$ mi.

Angular rates & path length

Problems 15, 18–19

(1)Wheel $\theta=128t-12t^{2}$. Find $\omega$ and $\alpha$ at $t=3$.

$\omega=128-24t$ ⇒ $\omega(3)=56$.

$\alpha=-24$.

$\omega=56$ rad/s; $\alpha=-24$ rad/s².

(2)Car $s=5t-t^{2}$ from $t=0$ to $t=3$. Total distance?

$v=5-2t=0$ at $t=2.5$; $s(0)=0$, $s(2.5)=6.25$, $s(3)=6$.

Distance $6.25+|6-6.25|=6.5$ miles.

$6.5$ miles.

21–22. Calculator / exact roots

Book 21. $s=t^{4}-3t^{2}+2t$: turns at $t=1$ and $t=(-1\pm\sqrt{3})/2$; left on $(-\infty,(-1-\sqrt{3})/2)$ and $((-1+\sqrt{3})/2,1)$.

Book 22. Paths $s=3t-t^{2}$ and $s=t^{3}-t^{2}+1$: same position near $0.35$ and $1.53$; same velocity at $t=\pm 1$; at meetings they move opposite ways.

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Mixed review

Motion mixed set

Rectilinear · distance · free fall · circular · supplementary

(1)Dropped from $1024$ ft ($v_0=0$). When does it hit, and with what speed?

$1024-16t^{2}=0$ ⇒ $t=8$.

$v=-32\cdot 8=-256$ ft/s; speed $256$ ft/s.

$t=8$ s; speed $256$ ft/s.

(2)From rest: $\theta=t^{3}/50-t$. Find $\theta,\omega,\alpha$ at $t=10$.

$\theta=10$ rad.

$\omega=3t^{2}/50-1=5$ rad/s.

$\alpha=6t/50=6/5$ rad/s².

$10$; $5$; $6/5$.

(3)For $s=t^{3}-6t^{2}+9t$ at $t=5/2$: position, direction, speed trend?

$s=5/8$; $v=-9/4$ (left); $a>0$ with $v<0$ ⇒ speed decreasing.

$s=5/8$; left; speed decreasing.

(4)For $s=t^{3}-6t^{2}+9t$ at $t=5/2$: position, direction, speed trend?

$s=5/8$; $v=-9/4$ (left); $a>0$ with $v<0$ ⇒ speed decreasing.

$s=5/8$; left; speed decreasing.

(5)For $s=t^{3}-6t^{2}+9t+4$: find $s$ and $a$ when $v=0$.

$t=1$: $s=8$, $a=-6$; $t=3$: $s=4$, $a=6$.

$(s,a)=(8,-6)$ and $(4,6)$.

(6)Does $s=t^{3}-3t^{2}+3t+3$ ever change direction?

$v=3(t-1)^{2}\ge 0$; double root at $t=1$ is not a turn.

No turns.

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Self-check

Quiz

Motion practice review

1 / 7

For s = ½t³ − 2t, a at t = 2 equals…

Pick an answer to see if you are right.

Further Reading

• 19. Rectilinear & Circular Motion — theory and explorers
• 20. Related Rates — next: ladder and dy/dt from dx/dt
• 14. Max and Min Values — extrema when v = 0
• 13. Law of the Mean — constants from equal derivatives
• 18b. Inverse Trig — Practice — prior chapter