# Thread: Am I Reading Too Much Into This?

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

[Problem] [Show Your Work]

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
[Problem] [Show Your Work]

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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