The hands of a clock are observed continuously from 12:45 pm onwards. They will be observed to point in the same direction some time between

Question

The hands of a clock are observed continuously from 12:45 pm onwards. They will be observed to point in the same direction some time between A).1:03 pm and 1:04 pm B).1:04 pm and 1:05 pm C).1:05 pm to 1:06 pm D).1:06 pm to 1:07 pm I don't understand how to solve the question. Plz give proper explanation.

Answer 1

If we define the angle $\theta$ as zero when the hand is vertical and increasing clockwise, the minute hand is at $\theta=0$ at $1:00$ and the hour hand is at $\theta=\frac \pi 6$. What is the angular speed of each hand? Write an equation for the angle of each hand $t$ minutes after $1:00$. Solve for when the angles are equal.

Answer 2

Okay, so you have change in angle per minute $ v = 5.5°/min$, starting angle ${\varphi }_{ 1}$and final angle ${\varphi }_{ 2} = 0°$. You need to find time that is needed to get from ${\varphi }_{ 1}$ to ${\varphi }_{ 2}$.

Starting angle ${\varphi}_{1}$ is equal to sum of angles ${\alpha}_{1}$ and ${\alpha}_{2}$

Since time is 12:45, minute hand is pointed to 9. Angle between 9 and 12 ( ${\alpha}_{1}$) is $3 \cdot 30°$

${\alpha}_{1} = 90°$

Hour hand is on $\frac { 3 }{ 4 } $ of an hour from 12.

${\alpha}_{2} = \frac{3}{4} \cdot 30° =22.5° $

Angle between the minute and hour hand is sum of $\alpha_{1}$ and $\alpha_{2}$.

${\varphi_{1}=112.5°}$

Therefore,

$t=\frac{{\varphi}_{1}-{\varphi}_{2}}{v}$

$t=\frac{112.5°}{5.5°/min}=20.454\quad min$

Your answer is then 12:45 + 20.454 $min$ = 13:05:27