Finding the Rate of distance between hands of clock

Question

First, I think I don't understand the problem which asks about the greatest rate of change in distance between the tips of the hands of clocks. Does it mean where the increasing of distance is the greatest in one minute? or what...? I used the laws of cosine $x^2=10^2+7^2-20*7\cos(\theta)$ where $x$ is the distance between hands and $\theta$ is the angle between hands.

A clock has a minute hand of $10cm$ long and an hour hand of $7cm$ long. The hands move at a constant rate (no jerking as in some clocks)

Problem : At what time is the distance between the tips of the hands changing at the greatest rate?

A. $12:00$

B. $12:05$

C. $12:10$

D. $12:15$

Answer 1

So from $12:00$ to $12:01$, the change in distance $\delta x=3-\sqrt{149-140\cos 5.5}\approx-0.11$

from $12:05$ to $12:06$, $\delta x=\sqrt{149-140\cos27.5}-\sqrt{149-140\cos33}\approx-0.64$

from $12:10$ to $12:11$, $\delta x \approx -0.66$,

and from $12:15$ to $12:16$, $\delta x\approx -0.57$.

The change was largest at $12:10$ in $1$ minute among the given choices.

Thus, the answer for the time of greatest rate in distance is $(C)\,12:10$.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \begin{array}{rcrcr} \mbox{'hour hand'} & \to & r_{\rm hh}\cos\pars{\omega_{\rm hh}t}\,\hat{x} & + & r_{\rm hh}\sin\pars{\omega_{\rm hh}t}\,\hat{y} \\ \mbox{'minute hand'} & \to & r_{\rm mh}\cos\pars{\omega_{\rm mh}t}\,\hat{x} & + & r_{\rm mh}\sin\pars{\omega_{\rm mh}t}\,\hat{y} \end{array}\,,\quad \left\vert% \begin{array}{rcr} r_{\rm hh} & \equiv & 7\ \mbox{cm.} \\[2mm] r_{\rm mh} & \equiv & 10\ \mbox{cm.} \\[2mm] \omega_{\rm hh} & \equiv & {2\pi \over 12\ \mbox{hours}} \\[2mm] \omega_{\rm mh} & \equiv & {2\pi \over 1\ \mbox{min.}} \end{array}\right. $$

$$ {\rm d}\pars{t} \equiv \sqrt{\bracks{% r_{\rm hh}\cos\pars{\omega_{\rm hh}t} - r_{\rm mh}\cos\pars{\omega_{\rm mh}t}}^{2} + \bracks{% r_{\rm hh}\sin\pars{\omega_{\rm hh}t} - r_{\rm mh}\sin\pars{\omega_{\rm mh}t}}^{2}} $$

With $\ds{{\rm D}\pars{t} \equiv {\dd\pars{t} \over r_{\rm mh}}}$ and $\ds{\mu \equiv {r_{\rm hh} \over r_{\rm mh}} < 1}$: $$ {\rm D}^{2}\pars{t} =1 + \mu^{2} - 2\mu\cos\pars{\omega t}\,,\qquad \omega \equiv \omega_{\rm mh} - \omega_{\rm hh} $$

\begin{align} 2{\rm D}\pars{t}\dot{\rm D}\pars{t} &= 2\mu\omega\sin\pars{\omega t} \\[1mm] 2\dot{\rm D}^{2}\pars{t} + 2{\rm D}\pars{t}\ddot{\rm D}\pars{t} &= 2\mu\omega^{2}\cos\pars{\omega t} \end{align}

The greatest rate occurs when $\ddot{{\rm D}}\pars{t} = 0$. Then, $$ 4\mu^{2}\omega^{2}\sin^{2}\pars{\omega t} =2{\rm D}^{2}\pars{t}\bracks{2\dot{\rm D}^{2}\pars{t}} =2{\rm D}^{2}\pars{t}\bracks{2\mu\omega^{2}\cos\pars{\omega t}} $$

$$ \mu\bracks{1 - \cos^{2}\pars{\omega t}} =\bracks{1 + \mu^{2} - 2\mu\cos\pars{\omega t}}\cos\pars{\omega t} =\pars{1 + \mu}^{2}\cos\pars{\omega t} -2\mu\cos^{2}\pars{\omega t} $$

$$ \mu\cos^{2}\pars{\omega t} - \pars{1 + \mu^{2}}\cos\pars{\omega t} + \mu = 0 \quad\imp\quad\cos\pars{\omega t} = \mu $$

$$ \omega t_{n} = \arccos\pars{\mu} + 2n\pi\,,\qquad n\ \in\ {\mathbb Z} $$ With these values of $t_{n}$ we'll have: $$\left\lbrace% \begin{array}{rcl} {\rm D}_{n}\pars{t} & = & \root{1 - \mu^{2}} \\ \dot{\rm D}_{n}\pars{t} & = & {\mu^{2} \over \root{1 - \mu^{2}}}\,\omega \end{array}\right. $$

Answer 3

Clearly, the net velocity vector will be larger when the hands' velocity vectors are perpendicular than when they are aligned, hence 12:15.