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Two trains leave towns 1212 kilometers apart at the same time and travel toward each other. One train travels 13 km/h slower than the other. If they meet in 4 hours, what is the rate of each train?

Answer :

[tex]\begin{gathered} \text{Let V}_1\text{ represent Velocity of the 1st train} \\ \text{Let V}_2\text{ represent Velocity of the 2nd train} \end{gathered}[/tex]

Thus,

[tex]\begin{gathered} V_1=V_1 \\ V_2=V_1-13 \end{gathered}[/tex]

The speed of an object/a body is calculated as:

[tex]\text{Speed, V=}\frac{dis\tan ce}{\text{time}}[/tex]

Thus, we have:

[tex]\begin{gathered} V_1=\frac{D_1}{t};V_2=\frac{D_2}{t} \\ \text{The two trains are 1212km apart} \\ \text{Thus, we have: D}_1+D_2=1212\operatorname{km} \\ \text{Time, t =4 hours} \end{gathered}[/tex][tex]\begin{gathered} V_1=\frac{D_1}{4} \\ \text{cross}-\text{multiply} \\ D_1=4V_1 \end{gathered}[/tex][tex]\begin{gathered} V_2=\frac{D_2}{4} \\ D_2=4_{}V_2 \\ \text{ Recall that V}_2=V_1-13 \\ \text{Thus, we have:} \\ D_2=4(V_1-13) \end{gathered}[/tex][tex]\begin{gathered} D_1+D_2=1212\operatorname{km} \\ 4V_1+4(V_1-13)=1212 \\ 4V_1+4V_1-52=1212 \\ 8V_1=1212+52 \\ 8V_1=1264 \\ V_1=\frac{1264}{8} \\ V_1=158\text{ km/hr} \end{gathered}[/tex]

To find the speed of the second train, we have:

[tex]\begin{gathered} V_2=V_1-13 \\ V_2=158-13 \\ V_2=145_{} \end{gathered}[/tex]

Hence, the rate of the first train is 158 km/hr and the rate of the second train is 145 km/hr

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