<p>They're starting to slip away from my memory. I'd greatly appreciate it if someone could briefly mention the formulas that address the following problems:</p>
<p>1- The average speed is required for two different speeds over the same distance.</p>
<p>2- The time is required for two different speeds over the same distance.</p>
<p>There are a couple more distance/speed/time formulas for different problems, that I just can't remember right now. If anyone could remind me of all these different forumlas, please. (Calling out to DrSteve, fogcity, xiggi, or any other to be math geniuses of this forum)</p>
<p>velocity is the first derivative of the position function
acceleration is the second derivative of the position function
jerk is the third derivative of the position function</p>
<p>This can be rewritten as 2 / (1/x + 1/y), which is the harmonic mean of the two speeds.</p>
<ol>
<li>We got that while we were doing the other one. The total time is **d(x+y) / xy<a href=“if%20a%20distance%20of%20d%20is%20covered%20at%20both%20speeds”>/b</a>.</li>
</ol>
<p>Example. A car goes for 40 mph for 20 miles, then 60 mph for 20 miles. What is the average speed and time?</p>
<p>Soln. d = 20, x = 40, and y = 60. Then the average speed is 2(40)(60) / (40+60) = 48 mph. The total time required is (20)(40+60) / (40)(60) = 20 / 24 = 50 minutes.</p>
<p>Example. On the way to the store, you travel at 30 mph. If you want your average speed on the road to be 40 mph, how fast do you have to be going on the way back?</p>
<p>Soln. v_av = 40 and x = 30. Plug in and solve for y:</p>
<p>40 = 2(30)(y) / (30 + y). y = 60 mph.</p>
<p>In my opinion, it’s not a good idea to try to memorize these formulas; they’re easy to forgot, misremember, or misuse. It’s much easier on your memory if you just remember d = vt, and use that several times in a problem.</p>
<p>I totally agree with conquerer7. For the distance/speed problems on the SAT all you need is: distance = speed x time, AND remember that this applies only over a distance where the speed is constant.</p>
<p>I encourage you to draw a precise diagram for distance/speed problems. Clearly identify each span where the speed is constant. Make the path very clear on the diagram, distinguishing direction clearly, as for example going from A to B, and returning from B to A. All this may sound trivial, but clear thinking is key in these kinds of problem and a diagram reinforces discipline.</p>
<p>Also, and again this may sound trivial, be very clear regarding what value is asked for. Note it on the diagram I’ve suggested. Often the list of choices include “answers” to related values. The test makers know that test takers may be careless and answer the “wrong” but related question.</p>
<p>If a question was asking for average speed of two speeds across multiple hours, what formula could we use?
As in, it took Sally 2 hours to go to and from her uncles house but she was going at a speed of 20 mph to and 15 mph from.</p>
<p>You’d use t_tot = d(x+y) / xy. Then 2 = d(35) / 300, and you can solve from there. It gives you d = 120/7 miles (hopefully that was what was being asked). </p>
<p>Still, it’s safer to just use d=vt. I’ve already forgotten the rest despite deriving them yesterday, and there’s no reason they can’t put a slight twist on a problem that would make all of the fancy formulas useless.</p>
<p>On the way there, d = 20(t). On the way back, d = 15(2-t). This is a system of linear equations you can quickly solve for d by eliminating t.</p>
<p>It took Sally 2 hours to go to and from her uncle’s house but she was going at a speed of 20 mph to and 15 mph from. What was her average speed.</p>
<p>The variant is taken from a totally different domain – instead of “physics”, “bakeshop economics”.</p>
<p>It took Sally 2 hours to sell all the cupcakes that she baked. She sold the first 1/2 of the cupcakes at the rate of 20 cupcakes per hour, and then after these were sold she sold the remaining 1/2 at the rate of 15 cupcakes per hour. What was her average selling rate (over the two hour period)?</p>
<p>The point of the variant is to emphasize that formulas don’t help and that the distance/speed problems are not physics problems. Simple definitions regarding “rate”, and the notion of “average” are all that matters. The SAT is a reasoning test after all – and with a very small number of exceptions no formulas are necessary. The formulas actually get in the way.</p>