Question

Question 8If it takes the same amount of time for a planet to move from point A to point B as itdoes for a planet to move from point C to point D, then what must a planet do interms of its speed in each sector?Sector 2Sector 116

Answer 1

We are given that to a planet takes the same amount of time to go through two different sectors. The speed of the planet (distance over time) when goes through sector AB is:

$v_{AB}=\frac{S_{AB}}{t}$

Where

$\begin{gathered} S_{AB}=\text{distanve from A to B} \\ t=\text{time} \end{gathered}$

And the speed from C to D is:

$v_{CD}=\frac{S_{CD}}{t}$

Where:

$\begin{gathered} S_{CD}=\text{distance from C to D} \\ t=\text{time} \end{gathered}$

We can solve for the time in each equation and we get:

$\begin{gathered} t=\frac{S_{AB}}{v_{AB}}_{} \\ \\ t=\frac{S_{CD}}{v_{CD}} \end{gathered}$

Since time is the same we can equate both equations:

$\frac{S_{AB}}{v_{AB}}=\frac{S_{CD}}{v_{CD}}_{}_{}$

Now we can cross multiply:

$v_{CD}S_{AB}=v_{AB}S_{CD}$

Solving for the speed CD:

$v_{CD}=\frac{v_{AB}S_{CD}}{S_{AB}}$

Since the distance CD is larger than the distance AB this means that:

$\frac{S_{CD}}{S_{AB}}>1$

Since the ratio is larger than 1, this means that speed CD is larger than speed AB:

$v_{CD}>v_{AB}$

This means that the planet will speed up at CD and slow down at AB.