Two rabbits start running towards each other, one from A to B and another from B to A. They cross each other after one hour and the first rabbit reaches B, hour before the second rabbit reaches A. If the distance between A and B is 50 km. what is the speed of the slower rabbit? 

(Work & Time)

প্রশ্নে বলা হচ্ছে যে, দুটি খরগোশের একটি A হতে B তে এবং অপরটি B হতে B তে যাচ্ছে। একঘণ্টা পর তারা পরষ্পর একে অপরকে অতিক্রম করলো। প্রথম খরগোশটি দ্বিতীয় খরগোশের চেয়ে ঘণ্টা বা 50 মিনিট পূর্বেই B স্থানে পৌছাল । A এবং B এই স্থান দুটির মধ্যবর্তী দূরত্ব 50 km হলে ধীরগতির খরগোশটির গতিবেগ কত ?
প্রথমেই প্রশ্ন অনুযায়ী নিচের মতো করে চিত্র আঁকুনঃ

Let, another rabbit takes x hours to go it's destination.

∴ One rabbit takes  $\left(x - \frac{5}{6}\right)$ hours to go it's destination.

As one rabbit reaches $\frac{5}{6}$ hours or 50 minutes before another rabbit, so one rabbit is faster and another one is slower.

Now, we assume that the speed of one rabbit is v₁ kmph and the speed of another rabbit is v₂ kmph.

$$\begin{gathered} As velocity = \frac{dis\tan ce}{time} \\ \therefore v_1 = \frac{50}{x - {\displaystyle \frac{5}{6}}}kmph . . . . . . . . . . . \left(i\right) \\ And v_2 = \frac{50}{x} kmph . . . . . . . . . . . .\left(ii\right) \\ As they are coming towards one another, their relative speed will be (v_1 + v_2) kmph. \\ Now, according to question \\ \Rightarrow \frac{50}{v_1 +v_2} = 1 \\ \Rightarrow \frac{50}{\left({\displaystyle \frac{50}{x-{\displaystyle \frac{5}{6}}}}\right)+\left({\displaystyle \frac{50}{x}}\right)}= 1 \\ \Rightarrow \frac{50}{50\left({\displaystyle \frac{1}{x-{\displaystyle \frac{5}{6}}}+\frac{1}{x}}\right)}= 1 \\ \Rightarrow \frac{1}{x-{\displaystyle \frac{5}{6}}} + \frac{1}{x} =1 \\ \Rightarrow \frac{1}{{\displaystyle \frac{6x-5}{6}}} + \frac{1}{x} =1 \\ \Rightarrow \frac{6}{{\displaystyle 6x-5}} + \frac{1}{x} =1 \\ \Rightarrow \frac{6x + 6x -5}{\left(6x-5\right)x}=1 \\ \Rightarrow \frac{12x -5}{6x^2-5x}= 1 \\ \Rightarrow 6x^2-5x-12x +5 = 0 \\ \Rightarrow 6x^2 -17x +5 =0 \\ \Rightarrow 6x^2 -15x-2x +5 = 0 \\ \Rightarrow 3x \left(2x-5\right) -1 \left(2x-5\right)= 0 \\ \therefore \left(2x-5\right) \left(3x-1\right) = 0 \\ So, the value of x will be either \frac{5}{2} or \frac{1}{3}. \left[But x = \frac{1}{3} is not acceptable\right] \\ Now, we will put value x = \frac{5}{2} in equation \left(ii\right) to get the speed of the slower rabbit. \\ \therefore v_2 = \frac{50}{{\displaystyle \frac{5}{2}}} = \frac{50}{1} \times \frac{2}{5} = 20 kmph \\ \therefore The speed of the slower rabbit is 20 kmph. \end{gathered}$$