redruM_0381 Posted May 3, 2006 Share Posted May 3, 2006 Im in a debate in a training class at work. Here is the given problem, WORD FOR WORD: "A train is travelling from boston to los angeles at 75mph and increasing its speed 5mph every hour. It left boston at 3am, what time will it get to LA." Assume the distance is 3000 miles, and you do figure the -3hr time difference. Seems simple enough. Others in class came up with answers and gave equations, etc. However, I feel this is unsolvable, because we do not know the rate of acceleration of the train during the hour, other than average over the hour. So, the train could accelerate instantly at each hour, it could be accelerating constantly, or it could take 2 minutes to get from 75-->80mph. A train that accelerates instantly at each hour would get further than the train that was constantly accelerating. Any math buffs around that can help me out? Quote Link to comment Share on other sites More sharing options...
OlderThanMe Posted May 3, 2006 Share Posted May 3, 2006 well for most problems like that that I have seen you assume the train instantly accelerates 5 mph... Quote Link to comment Share on other sites More sharing options...
zuhow Posted May 3, 2006 Share Posted May 3, 2006 I agree that's pretty crappy wording. It seems to me that the 5 mph per hour is a constant acceleration and the intial speed is 75 mph. simple "plug and chug" solution otherwise Quote Link to comment Share on other sites More sharing options...
Pop N Wood Posted May 3, 2006 Share Posted May 3, 2006 A train that accelerates instantly at each hour would get further than the train that was constantly accelerating. Other way around. 75 mph for the first hour, 80 mph for the second, etc. With a constant acceleration the speed would gradually increase for the first hour resulting in a shorter trip. Yeah, two ways it could be done. Set the accel = 5mph/hr then plug and chug, or just add up the distance traveled at a constant speed each hour until you get 3000 miles. Don't make it any more complicated than that. Quote Link to comment Share on other sites More sharing options...
redruM_0381 Posted May 3, 2006 Author Share Posted May 3, 2006 Other way around. 75 mph for the first hour, 80 mph for the second, etc. With a constant acceleration the speed would gradually increase for the first hour resulting in a shorter trip. I think you misunderstood me. A train that accelerates instantly from 75-->80mph at the first second of the hour will travel FURTHER in that hour, than the train that accelerates gently from 75-->80mph during the entire hour. Quote Link to comment Share on other sites More sharing options...
OlderThanMe Posted May 3, 2006 Share Posted May 3, 2006 hold on for a minute..I have a graph... Quote Link to comment Share on other sites More sharing options...
zuhow Posted May 3, 2006 Share Posted May 3, 2006 a train that accelerates 5mph instantly has an infinite acceleration. The number given is 5 mph per hour so that's a constant acceleration. s=v_0*t + 1/2*a*t^2 3000= 75t+ (1/2)*5* t^2 solve for t ==> ~22 hours Quote Link to comment Share on other sites More sharing options...
RB26powered74zcar Posted May 3, 2006 Share Posted May 3, 2006 a train that accelerates 5mph instantly has an infinite acceleration. The number given is 5 mph per hour so that's a constant acceleration. s=v_0*t + 1/2*a*t^2 3000= 75t+ (1/2)*5* t^2 solve for t ==> ~22 hours ^^^ That looks to me to be correct.... Quote Link to comment Share on other sites More sharing options...
OlderThanMe Posted May 3, 2006 Share Posted May 3, 2006 Quote Link to comment Share on other sites More sharing options...
redruM_0381 Posted May 3, 2006 Author Share Posted May 3, 2006 a train that accelerates 5mph instantly has an infinite acceleration. The number given is 5 mph per hour so that's a constant acceleration.s=v_0*t + 1/2*a*t^2 3000= 75t+ (1/2)*5* t^2 solve for t ==> ~22 hours I think you're right on the constant acceleration. However i get 23hours and change when I use the following equation to figure distance with constant acceleration: D = .5(Vf+Vi)T D=distance Vf = final Vel. Vi = Initial Vel. T=time Quote Link to comment Share on other sites More sharing options...
OlderThanMe Posted May 3, 2006 Share Posted May 3, 2006 see...speed X Time is distance...I think instant acceleration would be the fastest way. since there is moe time at a higher time than with constant acceleration. Quote Link to comment Share on other sites More sharing options...
zuhow Posted May 3, 2006 Share Posted May 3, 2006 see...speed X Time is distance...I think instant acceleration would be the fastest way. since there is moe time at a higher time than with constant acceleration. you're totally right in what you're saying, but that step change in velocity (going instantly from, say, 80 to 85 mph) is not a 5mph per hour acceleration- it's infinite (change in velocity/change in time, when change in time =0). that's the issue. If you had a step change in velocity of 5 mph every hour on the hour, you would certainly make it faster than if you had a constant acceleration of 5 mph/h- but that's not what the problem asked (or tried to ask). BTW The real time is closer to 23h- I think I calculated 22.7 or so. Quote Link to comment Share on other sites More sharing options...
OlderThanMe Posted May 3, 2006 Share Posted May 3, 2006 I just haven't heard that term much. I always heard "instant acceleration" from one speed to another. in the stepped graph there is no acceleration since there is no slope to the line. Ifinite change in speed..I got it. thanks... Quote Link to comment Share on other sites More sharing options...
zuhow Posted May 3, 2006 Share Posted May 3, 2006 I just haven't heard that term much. I always heard "instant acceleration" from one speed to another. in the stepped graph there is no acceleration since there is no slope to the line. Ifinite change in speed..I got it. thanks... close...in the graph the acceleration is infinite (vertical line) not zero (horizontal line). you got it that the acceleration is the slope of the velocity v. time graph , which is zero most of the time, but during the change it's infinite. make sense? Quote Link to comment Share on other sites More sharing options...
OlderThanMe Posted May 3, 2006 Share Posted May 3, 2006 well I would think it would not be infiinite now that I think about it...it would be indefinite...meaning that there is no time of acceleration. Y'all threw me a curveball with that one... Quote Link to comment Share on other sites More sharing options...
olie05 Posted May 3, 2006 Share Posted May 3, 2006 >> solve('75*x+(1/2)*5*x^2=3000') ans = -15+5*57^(1/2) -15-5*57^(1/2) >> -15+5*57^(1/2) ans = 22.7492 >> mmmm matlab7 as you can see, velocity is increasing, otherwise the graph would be one of a straight line. Quote Link to comment Share on other sites More sharing options...
jt1 Posted May 3, 2006 Share Posted May 3, 2006 That's just a poorly worded question that could have several correct answers depending on the assumptions needed to fill in the blanks. When I was in school....ahem....a few years back....the correct response would be to state your assumptions, then solve the problem based on those assumptions. You could also solve the problem for different assumptions and show the variations. Sometimes that might get an attaboy, sometimes not, depending on how the prof viewed the assumptions. JOhn Quote Link to comment Share on other sites More sharing options...
TheNeedForZ Posted May 3, 2006 Share Posted May 3, 2006 It's OK to assume "instant acceleration" math-wise since a real train can accelerate to 5mph faster in less than 10 secs. The difference between "instantly accelerate to 5 more mph" and "accelerate to 5 more mph in 10 secs" over a 3000mile trip is neglectable. To interprete the question word for word, it would mean the speed/time graph is stepped. Any object at speed is observed by their frame of reference to have slower time flow. So remember to throw that relativity thing in there. TECHNO BABBLE TIME! Quote Link to comment Share on other sites More sharing options...
OlderThanMe Posted May 3, 2006 Share Posted May 3, 2006 right on...but can a train a mile long accelerate 5 mph in 10 sec? noot likely. wait until you are at a train crossing and the train stops right on the crossing. It takes that train forever to take off again... Quote Link to comment Share on other sites More sharing options...
olie05 Posted May 3, 2006 Share Posted May 3, 2006 wow, 108 views in roughly 3hrs! i think i'd say there are some math nuts in here! Quote Link to comment Share on other sites More sharing options...
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