Two trains 16

Two trains cross each other in 14 seconds when moving in opposite directions, and in 3 minutes and 2 seconds when moving in the same direction. By what percentage is the speed of the faster train greater than that of the slower train?

Final Answer:

p =  16.6667 %

Step-by-step explanation:

$$\begin{gathered} t_1=14\text{s} \\ {t}_2=3\cdot 60+2=182\text{s} \\ {v}_1=r\cdot v_2 \\ {t}_1=\frac{s}{v_1+v_2} \\ {t}_2=\frac{s}{v_1-v_2} \\ {t}_1\cdot (v_1+v_2)=s \\ {t}_2\cdot (v_1-v_2)=s \\ {t}_1\cdot (v_1+v_2)=t_2\cdot (v_1-v_2) \\ {t}_1\cdot (r\cdot v_2+v_2)=t_2\cdot (r\cdot v_2-v_2) \\ {t}_1\cdot (r+1)=t_2\cdot (r-1) \\ 14\cdot (r+1)=182\cdot (r-1) \\ 168r=196 \\ r=\frac{196}{168}=1.16666667 \\ r=\frac{7}{6}\doteq 1.166667 \\ p=100\cdot (r-1)=100\cdot (1.1667-1)=16.6667\% \end{gathered}$$

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