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.
+36916 Distance d original rate r t = 5
d/r = 5 3/4 of the way is 3.75 hr 1.25 hr to go
gets delayed 1.5 hr for engine trouble
speeds up 10 and only arrives 1.25 late
so instead of 1.25 hr to finish ....when he speeds up
it only takes him 1 hour to finish the last 1/4 d
1/4 d /(r+10 ) = 1 hour
r = d/5
1/4 d / ( d/5+10) = 1
1/4 d = d/5 + 10
d = 200 km
