SOLUTION: Towns A and B are 220 km apart. A bus leaves Town A at 08 30 for Town B at a speed of 30 km/h. It arrives at Town B after stopping for 40 minutes at Town C which is 120 km from Tow
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-> SOLUTION: Towns A and B are 220 km apart. A bus leaves Town A at 08 30 for Town B at a speed of 30 km/h. It arrives at Town B after stopping for 40 minutes at Town C which is 120 km from Tow
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Question 437722: Towns A and B are 220 km apart. A bus leaves Town A at 08 30 for Town B at a speed of 30 km/h. It arrives at Town B after stopping for 40 minutes at Town C which is 120 km from Town A. A motorist starts from Town B at 09 00 and travels towards Town A at a speed of 40 km/h. Find graphically when and at what distance from Town A they will meet. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Towns A and B are 220 km apart. A bus leaves Town A at 08 30 for Town B at a speed of 30 km/h.
It arrives at Town B after stopping for 40 minutes at Town C which is 120 km from Town A.
A motorist starts from Town B at 09 00 and travels towards Town A at a speed of 40 km/h.
Find graphically when and at what distance from Town A they will meet.
:
The 40 min layover at C kind of screws things up but if B motorist arrives at C before A motorist, we can ignore it. See if that is true.
:
Town B is 100 km from C
:
Motorist A's time to C: 120/30 = 4 hrs, 8:30 + 4 = 12:30 PM arrival
Motorist B's time to C: 100/40 = 2.5 hrs, 9:00 + 2:30 = 11:30 AM arrival
:
Motorist B departs C at 11:30
Motorist A departs A at 8:30
:
Let t = A's travel time till they meet
then
(t-3) = B's' travel time, ///I corrected a mistake here
:
A distance expression for each motorist, referenced to City A
A's dist = 30t; (red line)
B'a dist = 120 - 40(t-3); green line
:
Intersect: t = 3.43 hrs, dist ~ 103 mi from A
:
3.43 hrs = 3 hrs 26 min: 8:30 + 3:26 = 11:56 AM they meet 103 mi from A
