A driver set out from A at a certain rate and would have reached B in 5 hours if he had continue at this rate,However ,when we had gone three-fourth of the way he had motor trouble which delayed him for an hour and a half.During the remainder of the journey he increased his speed by 10 km an hour and reached B one hour and 15 minutes behind schedule.Find the distance from A to B.
Let his average speed BEFORE motor problem =S
The distance between A and B =D
The distance between A and where he had motor problem =3/4D
The remaining distance to B =1/4D
3.75(hours) + 1.5(hours) + [1/4D/(S+10)] =6.25, solve for D, S
D = 200 km - distance between A and B
S =40 kp/h before motor problem.
40+10 = 50 kp/h - his speed the last 1/4D