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Given a distance, two different speeds and time how to determine how far was traveled on each speed.

For example, if 100m has been traveled in 5 seconds and there was the option to travel the distance by bike and on foot. Speed on bike is 40m/s and speed on foot is 10m/s, how far did the person travel by bike and how much on foot? (Yes values are not actual representation of the real world)

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traveled distance=D=100

time taken=t=10

t=time on foot+time on bike=tf+tb

D=10*time on foot+40*time on bike=10tf+40tb

100=10tf+40tb => tf=10-4tb

edit: time taken was actually 5, not 10, but the idea is the same

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  • $\begingroup$ I got this far as well, but the problem is tf nor tb is known $\endgroup$ – user1435395 Sep 27 '13 at 8:01
  • $\begingroup$ Well tf+tb=5, tf=10-4tb. substituting: 10-4tb+tb=5 -> -3tb=-5 ->tb=5/3 -> tf=10-4*(5/3)=1/3 $\endgroup$ – user2520938 Sep 27 '13 at 8:06
  • $\begingroup$ Totally forgot about the substitution, let's just say it was still early $\endgroup$ – user1435395 Sep 27 '13 at 8:28
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Let $t_1$ denote the time travelled on foot and $t_2$ the time travelled by bike. Then we have on one side $$ t_1 + t_2 = 5$$ on the other hand, if we consider the travelled distance $$ 10t_1 + 40t_2 = 100. $$ Can you do it from here?

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  • $\begingroup$ I got this far as well, but the problem is t1 nor t2 is known $\endgroup$ – user1435395 Sep 27 '13 at 8:00
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You say that you've gotten to the point where you have two equations and two variables - so what's the problem? Do you need help with the algebra? Solve for t1 in one equation in terms of t2, then substitute that in the second equation to solve for t2.

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  • $\begingroup$ Totally forgot about the substitution, let's just say it was still early $\endgroup$ – user1435395 Sep 27 '13 at 8:29

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