So i know some of you here are pretty good in math, help me out please: at a high-school cross-country meet, jarryd jogs 8 mph for the first part of the race, then increased his speed to 12 mph for the second part. If the race was 21 mi long and Jared finished in 2 hr, how far did he jog at the faster pace? any help would be great, kthks.
I'm not gonna solve it for you but think of it this way, when he's running 12 mph every 20 minutes he has traveled 4 miles. If he finishes in 2 hours, then he must have at least spent an entire hour running at 12 mph. Hope that helps.
I solved it like this: (answer is in white in case anybody else is trying to solve it) If he traveled 1 hour at each speed, he would have traveled 20 miles total (8+12mph). Therefore he traveled a little longer at 12 mph than he did at 8 mph. you could also say that he goes 1 mile every 7.5 minutes at 8 miles an hour, and every 5 minutes at 12 miles an hour. In terms of an equation, it would be 7.5x + 5y = 120. Schimmel was right in saying that at least half of the race must have been at 12 mph, so we can change the equation to 7.5x + 5y = 60 (knowing that half of the race was already run at 12mph). After a little guessing and checking, I determined that he ran 1 hr 15 minutes at 12 mph, and 45 minutes at 8 mph. You can check this through the equation (miles run @ 8mph * minutes/mile @ 8mph) + (miles run @ 12mph * minutes/mile @ 12mph) = (total minutes ran), or (6 * 7.5) + (15 * 5) = 120.
Break up the problem into two parts: the distance he jogged at 8mph and the distance he jogged at 12mph. Breaking it up and looking at them separately will give you two linear velocity questions which you can solve.
Sorry, but wouldn't that be a kind of a logic reasoning to get to that? Thing im worried about is that it should follow some form of D=Rt as a uniform motion exercise as the text book says
Find an answer like that and I'll give you 5 bucks... What? I hate math so much i want it to be destroyed.
@piracer. It's still based off of d = rt. But you're extending d=rt a little into d = rt + r't' [the ' denotes a different set of rate and time]. And if there is a fixed total time like this, then T = t + t' (or 2 = t + t'), which can be rearranged into t' = 2 - t. Substitution. d = rt + r'(2-t). Algebra...it makes life simpler... Hope that makes sense.
This topic was a lot more interesting when I thought the title was meth problems, 8 years here and everyones still doing their schoolwork together
Erm well i can come to an answer that 45min was done at 8mph and the rest at 12mph (since that totals up nicely). the issue is i still don't quite know how to flesh that out as a formula and not just magically jot those numbers down