Problema Solution
Two cyclists leave a city at the same time, one going east and the other going west. The west bound cyclist bikes at 3 mph faster than the east bound cyclist. After 5 hours they are 195 miles apart. How fast is each cyclist going?
Answer provided by our tutors
Let
v1 is the speed of the west bound cyclist
v2 is the speed of the east bound cyclist
Then v1 = v2 + 3.
After time of t = 5 hours
the west bound cyclist has passed S1 = v1*t = 5*v1 miles
the east bound cyclist has passed S2 = v2*t = 5*v2 miles
Because they are cycling in opposite directions (one to west the other to east) the passed distance is equal to the shortest possible distance between them that is 195 miles. Thus we have
S1 + S2 = 195
5*v1 + 5*v2 = 195
divide both sides by 5
v1 + v2 = 195/5
v1 + v2 = 39
If we replace v1 = v2 + 3 in the last equation we get
v2 + 3 + v2 = 39
2*v2 + 3 = 39
2*v2 = 39 - 3
2*v2 = 36
v2 = 36/2
v2 = 13 mph
V1 = 13 + 3 = 16 mph
The west bound cyclist bikes at 16 mph while the east bound cyclist at 13 mph.