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

Answer:

$v_{1}$ = 5 km/hr

$v_{2}$= 4.5 km/r

Step-by-step 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