<p>[2<em>Speed1</em>Speed2] / [speed1 + Speed2]</p>
<p>Why does it work for 1 hour speed problems?</p>
<p>He posted the derivation somewhere. I got a question like this one in the Oct SAT, saved me a LOT of time.</p>
<p>someone want to link me to such a derivation ?</p>
<p>What is a “1 hour speed problem”?</p>
<p>Yeah… what is this?</p>
<p>A girl rides her bicycle to school at an average speed of 8 mph. She returns to her house using the same
route at an average speed of 12 mph. If the round trip took 1 hour, how many miles is the round trip.
A. 8
B. 9 3/5
C. 10
D. 11 1/5
E. 12 </p>
<p>It’s Xiggi’s formula to solve a problem like that ^</p>
<p>I don’t know how he found/derived it though.</p>
<p>I thought it only worked for average speed? not total distance</p>
<p>EDIT: wait nevermind I see why it would work for 1 hr problems b/c u just multiply the average speed by 1. LOL</p>
<p>Speed*Time = Distance</p>
<p>The distance to a place and from a place should be the same, and since distance is the product of speed and time:</p>
<p>speed1<em>time1 = speed2</em>time2</p>
<p>Because time1 + time2 = 1, time1 = 1 - time2, so</p>
<p>speed1<em>(1-time2) = speed2</em>time2
speed1 - speed1<em>time2 = speed2</em>time2
speed1 = speed1<em>time2 + speed2</em>time2
speed1 = time2(speed1 + speed2)
time2 = speed1/(speed1 + speed2)</p>
<p>Because both distances are equal, the total distance will be twice the length of either half of the trip. Using the second half of the trip, we have</p>
<p>distance/2 = speed2<em>time2
distance/2 = speed2</em>[speed1/(speed1 + speed2)]</p>
<p>And finally, we have</p>
<p>distance = 2<em>speed1</em>speed2/(speed1 + speed2)</p>
<p>which is the desired equality.</p>
<p>There’s probably a much simpler way to derive that, but this worked.</p>
<p>^ clever. ~</p>
<p>Expanding on my previous post, if ‘time’ represents the total time of the round trip, the total distance is given by 2<em>time</em>speed1*speed2/(speed1 + speed2).</p>