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January 11th, 2017, 08:54 PM  #1 
Member Joined: May 2015 From: Australia Posts: 36 Thanks: 5  Year 10 Motion problem
Question: A motor cyclist makes a trip of 500km. If he had increased his speed by 10km/h, he could have covered 600km in the same time. What was his original speed? Answer: 50km/h I tried solving it by using d=st st + t(s+10) = 500 + 600 st + st + 10t = 1100 2st + 10t = 1100 I think I'm doing it wrong though. Could someone help me? Last edited by skipjack; January 12th, 2017 at 04:29 AM. 
January 11th, 2017, 10:06 PM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 982 Thanks: 347 Math Focus: Yet to find out. 
The key here is that in both cases, the time is the same. So for the first case, we have the following, $\displaystyle 500 = \frac{1}{2}v t$ And so, $\displaystyle t = \frac{1000}{v}$. Now if we follow a similar procedure, this time having a known value for t, we have, $\displaystyle 600 = \frac{1}{2} (v + 10) * \frac{1000}{v}$ $\displaystyle 1200 = \frac{1000 \cancel{v}}{\cancel{v}} + \frac{10000}{v}$ $\displaystyle (1200  1000)v = 10000$ $\displaystyle \therefore v = 50km/h.$ Note that I have been sloppy with labeling my units here. Always remember what units you are working with! Last edited by skipjack; January 12th, 2017 at 03:51 AM. 
January 12th, 2017, 04:27 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 16,782 Thanks: 1238  As distance increase = (speed increase) * time, time = 100km/(10km/h) = 10h, and so the original speed = 500km/(10h) = 50km/h. If you prefer to use your variables, you can write that as shown below. Subtracting st = 500km from (s + 10km/h)t = 600km gives (10km/h)t = 100km, and so t = 10h. Hence s = 500km/t = 50km/h. 
January 12th, 2017, 05:51 AM  #4 
Member Joined: May 2015 From: Australia Posts: 36 Thanks: 5 
I understand your solution, but in my textbook there is an example preceding the question. In the example, it adds the variables instead of subtracting it. Is there a way to do it similar to the example method or not?

January 12th, 2017, 07:10 AM  #5 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 1,928 Thanks: 628 Math Focus: Physics, mathematical modelling, numerical and computational solutions  That's a different problem entirely, so it has a different solution. Don't compare them in terms of "adding things" or "subtracting things" because you shouldn't work like that. If you stick to the formulas, you'll be fine.


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