Question

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

Answer 1

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

Thus,

$\begin{gathered} V_1=V_1 \\ V_2=V_1-13 \end{gathered}$

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

$\text{Speed, V=}\frac{dis\tan ce}{\text{time}}$

Thus, we have:

$\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}$$\begin{gathered} V_1=\frac{D_1}{4} \\ \text{cross}-\text{multiply} \\ D_1=4V_1 \end{gathered}$$\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}$$\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}$

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

$\begin{gathered} V_2=V_1-13 \\ V_2=158-13 \\ V_2=145_{} \end{gathered}$

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