Question

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

Answer 1

`v_(1)` = 5 km/hr

`v_(2)`= 4.5 km/r

Explanation:

Distance of two tourists can be represented by
`d_(1)` and
`d_(2)`

Therefore, the total distance would be

`d_(1)` +
`d_(2)` =38 ----> (eq1)

->if the first one covered 2 km more than the second

`d_(1)` =
`d_(2)` +2 (substituting this in above equation)

We will have,

(
`d_(2)` +2)+
`d_(2)` =38

2
`d_(2)`+2=38

2
`d_(2)`=36

`d_(2)` = 36/2

`d_(2)`=18

plugging value of
`d_(2)` in eq(1)

eq(1)=>

`d_(1)` +
`d_(2)` = 38

`d_(1)` + 18 =38

`d_(1)` = 20 km

the distance of each of the tourists

`d_(1)` =20km and
`d_(2)` =18km

As they both traveled for 4 hrs

t=4hr

speed can be defined as distance per unit time i.e

speed 'v' = distance 'd' / time 't'

Let
`v_(1)` and
`v_(2)` represent speed of each of the tourists

Therefore,

`v_(1)` =
`d_(1)` /t => 20/4

`v_(1)` = 5 km/hr

`v_(2)`=
`d_(2)` /t => 18/4

`v_(2)`= 4.5 km/r