<p>Doing a physics assignment on webassign, and I've gotten everything but this last question.</p>
<p>When I looked at it, I thought "Wow, basic algebra," entered answer, wrong. It's:</p>
<p>"A race car driver must average 203.0 km/h over the course of a time trial lasting ten laps. If the first nine laps were done at 200.0 km/h what average speed must be maintained for the last lap? (Answer to the nearest km/h)"</p>
<p>First I did basic stuff for an average. 203=[200(9)+x]/10 , x being the speed for the last lap and 10 being the total number of laps. I got 230 km/h, but it says that is wrong.</p>
<p>Then I used a formula our teacher gave us for average velocity. (average velocity=(initial velocity+velocity)/2). For that I tried 203=(200+v)/2 and it gave me 206 km/h, which didn't sound right, and it wasn't.</p>
<p>Last I checked, my teacher was a bit confused when the website told him he got it wrong too. What am I missing here?</p>
<p>your teacher got it wrong? Maybe the site is messed up?</p>
<p>The weird thing is, some girl in the class said she asked her dad and found some way to do it. Nobody followed what she did, and nobody had time to ask because it was the end of class.</p>
<p>Careful. You're trying to find the average over a period of TIME, not distance. Since you'll spend <em>less time</em> doing the last lap if you go faster, you can't just use 203=(200*9+x)/10. </p>
<p><em>sighs</em> your teacher should have been able to explain that; it's a common confusion.</p>
<p>Try it this way: suppose a lap has L kilometers. If the driver maintains a speed of x kph during the last lap, his time for all 10 laps = time for 1st 9 laps + time for last lap
= 9(L/200) + 1(L/x)
This has to be the same as the time he would get if he had done all 10 laps at 203 kph i.e. 10(L/203)</p>
<p>So, 9(L/200) + (L/x) = 10(L/203)
(1/x) = 10/203 - 9/200 = 0.04926 - 0.045 = 0.00426
and x = 1/0.00426 = 234.74 kph or 235 kph, rounding off to the nearest integer value</p>
<p>Ah, I see now. Thank you very much, especially optimizerdad.</p>