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.

Guest Jul 14, 2020

#1**+1 **

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

ElectricPavlov Jul 14, 2020