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Two cyclists A and B starts from the junction of two roads making a right angle. The ratio of velocities $3:4$. Find the ratio of the rate at which two cyclists are separating with the velocity of A.

I could not approach the problem much. Please help.

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  • $\begingroup$ What do you mean by "the ratio of the rate at which two cyclists are separating with the velocity of a". So is "a" defined or... $\endgroup$
    – Macrophage
    Dec 30 '17 at 11:14
  • $\begingroup$ @Macrophage I guess you are right. Maybe the a is a typo that crept in the book. Would removing the a clarify the problem? $\endgroup$
    – Swadhin
    Dec 30 '17 at 12:34
  • $\begingroup$ It's just wierd that the ratio of two cyclists' speeds is given, so I'm not sure what ratio the problem is asking us to look for. $\endgroup$
    – Macrophage
    Dec 30 '17 at 12:38
  • $\begingroup$ @Macrophage Does it make much sense now? $\endgroup$
    – Swadhin
    Dec 30 '17 at 13:32
  • $\begingroup$ Check it out. :P $\endgroup$
    – Macrophage
    Dec 30 '17 at 15:15
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Since the two cyclists have a $3:4$ ratio of velocities, we assume they are $3v, 4v $ correspondingly.

Then, the distance of cyclist A from the intersection is given by $d_1=3vt$. In the same way, the distance of cyclist B from the intersection is given by $d_2=4vt$. By Pythagorean's theorem we know the distance between the two cyclists is $d=\sqrt{d_1^2+d_2^2}=5vt$.

Thus, the rate at which two cyclists are separating can be found by taking derivative of $d$ with respect to time $t$. $\dfrac{d}{dt}5vt=5v$

Therefore, the ratio ratio of the rate at which two cyclists are separating with the velocity of A should be $\frac{5v}{3v}=\frac{5}{3}$. You got it!

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    $\begingroup$ [+1] You got it... Speaking in terms of speed vectors (with a figure) would even be more convincing... $\endgroup$
    – Jean Marie
    Jan 2 '18 at 1:53
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    $\begingroup$ @JeanMarie Yeah, you are right. But since this is a calculus question, I just tried to show some calculus work by differentiating displacement to get velocity, though directly using velocity will make things even easier :P $\endgroup$
    – Macrophage
    Jan 2 '18 at 2:00
  • $\begingroup$ I agree : this kind of question can be asked to people that aren't already familiarized with vectors. $\endgroup$
    – Jean Marie
    Jan 2 '18 at 2:02

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