# Two tourists left two towns simultaneously, the distance between which is 38 km, and met in 4 hours. What was the speed of each of the tourists, if the first one covered 2 km more than the second one before they met?

Apr 27, 2018

$\implies {v}_{1} = 5 \text{km"/"hr}$

$\implies {v}_{2} = 4.5 \text{km"/"hr}$

#### Explanation:

Let ${d}_{1}$ and ${d}_{2}$ be the distances traveled in $\text{km}$ by each of the tourists.

We can write the total distance traveled as:

${d}_{\text{tot}} = {d}_{1} + {d}_{2} = 38$

We are told directly that the first tourist travels more than the second tourist:

${d}_{1} = {d}_{2} + 2$

We use these two equations to find the distance each tourist covered.

$\left({d}_{2} + 2\right) + {d}_{2} = 38$

$2 {d}_{2} + 2 = 38$

${d}_{2} + 1 = 19$

${d}_{2} = 18$

Substituting back to find ${d}_{1}$:

${d}_{1} = {d}_{2} + 2 = 18 + 2 = 20$

So we have found ${d}_{1} = 20 \text{ km}$ and ${d}_{2} = 18 \text{ km}$.

We know that each tourist traveled for $t = 4 \text{ hr}$. Velocity is defined as distance per unit time, so we can compute the velocities using the time and distances we found earlier.

${v}_{1} = {d}_{1} / t = \frac{20}{4} = \textcolor{b l u e}{5 \text{km"/"hr}}$

${v}_{2} = {d}_{2} / t = \frac{18}{4} = \textcolor{b l u e}{4.5 \text{km"/"hr}}$