Answer:
The answer is below
Step-by-step explanation:
The question is not complete, but let me solve a similar question:
Three trains one eastbound, one westbound, and one northbound leave a city at the same time. The speed of the northbound train is 20 miles per hour greater than the speed of the eastbound train. After 6 hours, the distance between the westbound train and the eastbound train is 420 miles. Twice the speed of the westbound train is 30 miles per hour more than the speed of the northbound train. Find the speeds of the three trains.
Solution:
Speed is the time rate of change of distance. It is the ratio of distance to time taken. Speed is given by:
Speed = distance / time
Let x represent the speed of the eastbound train, y that of the westbound train and z that of the northbound train. Hence:
z = x + 20
x - z = -20 (1)
After 6 hours, the distance between the westbound train and the eastbound train is 420 miles. Hence:
6 hours * x + (6 hours * y) = 420
6x + 6y = 420
x + y = 70 (2)
Lastly:
2y = z + 30
2y - z = 30 (3)
Solving equations (1), (2) and (3) simultaneously gives:
x = 30 miles per hour, y = 40 miles per hour, z = 50 miles per hour.
Hence the speed of the eastbound train is 30 miles per hour, speed of the westbound train is 40 miles per hour and the speed of the northbound train is 50 miles per hour.
