# Relative speed of two cyclists starting from a junction of two roads making a right angle with a certain fixed velocities ratio

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.

• 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... Dec 30 '17 at 11:14
• @Macrophage I guess you are right. Maybe the a is a typo that crept in the book. Would removing the a clarify the problem? Dec 30 '17 at 12:34
• 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. Dec 30 '17 at 12:38
• @Macrophage Does it make much sense now? Dec 30 '17 at 13:32
• Check it out. :P Dec 30 '17 at 15:15

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!