Answer:
2) 15 hours
3) 1080 miles
Explanations:
Let the distance traveled by the first car be x
Let the distance traveled by the second car be y
The total distance = 1575 miles
x + y = 1575..........................(1)
Speed of the first car = 55 mph
Speed of the second car = 50 mph
Time = Distance/speed
Time spent by the first car = x/55
Time spent by the second car = y/50
Both cars meet after spending the same time on the road
x/55 = y/50
50x = 55y....................(2)
From equation (1)
x = 1575 - y..........(3)
Substitute equation (3) into equation (2)
50(1575 - y) = 55y
78750 - 50y = 55y
50y + 55y = 78750
105y = 78750
y = 78750/105
y = 750
Time spent = y/50
Time spent = 750/50
Time spent = 15 hours
The two cars meet after 15 hours
3)
There are 9 hours from 9AM to 6 PM
Time spent by the first car = x/55
9 = x/55
x = 9(55)
x = 495 miles
Since the total distance is 1575
The distance left for the car to reach Albuquerque = 1575 - 495
The distance left for the car to reach Albuquerque = 1080 miles
If the Cleveland to Albuquerque car (from problem 2) left at 9 A.M, it will be 1080 miles far from Albuquerque at 6 PM