Ok so I have this homework problem. I dont want to give out the information because i want to put in the values myself, but basically I have one object moving at said speed. Said time later, another object leaves the same location as the object 1 and said speed faster. At some time later, object 2 is only said distance away from object 1. I have to find both their speeds. If this is too vague then i can re-write the question. Anybody got any tips?
Physics: Uniform Motion
5
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algebra-precalculus
physics
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0You should move this question to http://physics.stackexchange.com – 2011-09-13
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0ok thanks for the commentary – 2011-09-13
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0Use $\text{distance}=\text{rate}\times(\text{time}-\text{delay})$ to set up an equation. – 2011-09-13
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0ok, so by rate your referring to the time passed over the average speed right? – 2011-09-13
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1Your post does not make absolutely clear what information is given. Let $v$ be the speed of car 1. Let $v+k$ be the speed of car 2. Let $t_2$ be the time car 2 waits before taking off, and let $t$ be time it has travelled. Let $d$ be the distance it is still "behind." When car 2 started, it was $vt_2$ behind. It has gained $kt$, so it is $vt_2-kt$ behind. So $vt_2-kt=d$. If we know $k$, $t_2$, $t$, $d$, we can find $v$. [But if *distance* only is given, maybe car 2 is now *ahead*, then $kt-vt_2=d$. Whether this should be looked at depends on wording details of the question.] – 2011-09-13
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0Ok well car2 is behind car 1 by said distance(10 miles). I have to figure out both cars speeds – 2011-09-13
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0the only things that were given to me were the time car1 took off, the time it took car2 to take off from car1 starting point, and the amount of time it took for car2 to be said distance behind car 1. and the difference in speed in both cars. (car 1 going at (lets call it x) and car 2 going at (lets call it x+z) – 2011-09-13
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0You should be able to use the formula I mentioned in the comment. From what you say, after a while object 2 is 10 miles behind, so $d=10$. You are told my $k$ is $50$ (object 2 is 50 mph faster). The wording of your problem proably lets you find $t_2$ (how long object 2 waited) and $t$. So $vt_2-kt=d$, therefore $v=(d+kt)/t_2$. All numbers on the right-hand side are presumably known. – 2011-09-14
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1@Ronnie: It would have been better if you had given the problem fully, with all numbers, and requested that people not solve the problem for you, only give some idea of how to proceed. – 2011-09-14
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0Distance traveled by the first object moving at velocity $v$ and departing at time $t=0$ $$s_{1}=vt.$$ Distance traveled by the second object moving at velocity $v+c$ and departing with a delay $t_{0}$ with respect to the first object $$s_{2}=(v+c)(t-t_{0})\qquad t\geq t_{0}.$$ What is the distance $d$ between both objects at instant $t=t_{0}+t_{1}$ ($t_{1}$ after $t_{0}$)? $$d=s_{1}-s_{2}.$$ – 2011-09-14
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0Ok Andre. A train leaves chicago at 12. 2 hours later another train leaves chicago at a speed 50 miles an hour faster than train A. After an hour, train B is 10 miles behind train A. what are both their speeds – 2011-09-14
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0i was hesitant about writing the actual problem because someone might close it down – 2011-09-14
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0If I used your formula correctly, I got 30 miles for train 1 and 80 for train 2 – 2011-09-14
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1Yeah, let's credit Ronnie for an ingenious way of making sure that he will only get hints (ok, may be a formula) as opposed to a solution. Wouldn't mind seeing more of that actually! – 2011-09-14