Question

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

Answer 1

Answer:

⇒5 km/hr

⇒4.5 km/hr

Explanation:

Let d1 and d2 be the distances traveled in km by each of the tourists. We can write the total distance traveled as:

d1+d2=38

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

d1=d2+2

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

(d2+2)+d2=38

2d+2=38

d2+1=19

d2=18

Substituting back to find d1:

d\1=d2+2=18+2=20

So we have found

d1=20km and d2=18km

We know that each tourist traveled for

t=4hr

Velocity is defined as distance per unit time, so we can compute the velocities using the time and distances we found earlier.

v1=d1

t=204=5km/hr

v2=d2t=184=4.5km/hr

Answer 2

Let the relative speed will be given by:
38/4=9.5 km/h

Let the speed of the first tourist be x km/h
the speed of the second one is (9.5-x) km/h
distance covered by the first one is:
4*x+2=4x+2 km
distance covered by the second one is:
(9.5-x)4 km
=(38+4x) km
total distance will be given by:
(38+4x)+(4x+2)=38
40+8x=38
8x=-2
x=-1/4 km/h
this mean that he was traveling in the opposite direct thus the  speed was 1/4 km/k
The speed of the second one will be:
9.5-1/4=9.25 km/h