1. ## Am I Reading Too Much Into This?

The average distance between the earth and the sun is 93 million miles. Using the velocity of light, find an approximate value for the amount of time it takes a light ray to travel from the sun to the earth.

(Velocity of Light = 186,000 miles per second)

2. Originally Posted by ITTTechstudent

The average distance between the earth and the sun is 93 million miles. Using the velocity of light, find an approximate value for the amount of time it takes a light ray to travel from the sun to the earth.

(Velocity of Light = 186,000 miles per second)
Um, I could be reading too little into it, but I'd say:

$average~time = \frac{93,000,000 ~miles}{186,000 ~\frac {miles}{seconds}}$
miles cancel out, seconds go to numerator, and 93000/186=500 so the average time is 500 seconds.

3. hmm... the answer u ended up with seems more rational than mine...i had no where near 500 secs lol...so that could be right..

thanks!

4. Originally Posted by ITTTechstudent
hmm... the answer u ended up with seems more rational than mine...i had no where near 500 secs lol...so that could be right..

thanks!
Some calculus might make it a bit clearer, if you know that the average value of a function $f$ on an interval $[a,\,b]$ is $\overline f = \frac1{b - a}\int_a^bf(x)\,dx$.

Suppose at a given time $t$ the Sun is a distance $s(t)$ miles from the Earth, and further suppose that the Earth revolves around the Sun exactly once between times $t = a$ and $t = b$.

Then any light ray hitting the Earth at a time $t$ will have traveled for $\frac{s(t)}{186\cdot10^3}$ seconds. As $s$ should be a continuous function of $t,$ we may determine the average time to be

$\overline t = \frac1{b - a}\int_a^b\frac{s(t)}{186\cdot10^3}\,dt$

$=\frac1{\left(186\cdot10^3\right)(b - a)}\int_a^b s(t)\,dt$

$=\frac1{186\cdot10^3}\left(\frac{\int_a^b s(t)\,dt}{b - a}\right)$

$=\frac1{186\cdot10^3}\cdot\overline s$

$=\frac{\overline s}{186\cdot10^3}$

$=\frac{93\cdot10^6}{186\cdot10^3} = \frac{10^3}2 = 500$

Carrying the units through, the answer is in seconds, and this agrees with angel.white's solution.

5. Originally Posted by Reckoner
Some calculus might make it a bit clearer, if you know that the average value of a function $f$ on an interval $[a,\,b]$ is $\overline f = \frac1{b - a}\int_a^bf(x)\,dx$.

Suppose at a given time $t$ the Sun is a distance $s(t)$ miles from the Earth, and further suppose that the Earth revolves around the Sun exactly once between times $t = a$ and $t = b$.

Then any light ray hitting the Earth at a time $t$ will have traveled for $\frac{s(t)}{186\cdot10^3}$ seconds. As $s$ should be a continuous function of $t,$ we may determine the average time to be

$\overline t = \frac1{b - a}\int_a^b\frac{s(t)}{186\cdot10^3}\,dt$

$=\frac1{\left(186\cdot10^3\right)(b - a)}\int_a^b s(t)\,dt$

$=\frac1{186\cdot10^3}\left(\frac{\int_a^b s(t)\,dt}{b - a}\right)$

$=\frac1{186\cdot10^3}\cdot\overline s$

$=\frac{\overline s}{186\cdot10^3}$

$=\frac{93\cdot10^6}{186\cdot10^3} = \frac{10^3}2 = 500$

Carrying the units through, the answer is in seconds, and this agrees with angel.white's solution.
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