(1st) take the 1/2 hour stop out of the Late Times because the Stop Time does NOT Contribute to the Distance and Find the Usual Pre-Accident Speed, which we will call R
Focusing just on the 90km Difference in each Scenario:
Scenario 1:
---------------{A} ----------------Rest @(3/4)R----------------> 3 hours Late
Scenario 2:
-----------------------------------{A - 90 km later} -------------Rest @ (3/4)R-----------> 2.5 hours Late
Logic:
in Scenario 2, he was driving an Extra 90 km @ Usual Speed R ------> and was only 2.5 hours late
in Scenario 1, during this SAME 90 km stretch, because the accident occurred earlier, he was driving @ (3/4)R ------> and was 3 hours late
Over this Constant Distance of 90 km, the Speed is Inversely Proportional to the Time taken
Speed driven over this 90 km in Scenario 2 is R --------> in Scenario 1 it is (3/4)R, thus Speed DECREASED by (1/4)
therefore, the Time taken to cover the 90 km from Scenario 2 --------> to Scenario 1 INCREASED by (1/3)
because the accident occurred 90km EARLIER in Scenario 1, the extra time he was Late by in Scenario 1 was:
(3 hours) - (2.5 hours) = 1/2 hour
in Summary, over this 90 km Distance:
When the Speed DECREASED by (1/4) ---------> the Time INCREASED by (1/3) and this Increase in Time corresponds to an Extra + (1/2) hour of Lateness
Let the Usual Time needed to cover 90km
The Usual Speed of R = T
T + (1/3)T = T + 1/2 hour
T = (3/2) hour
thus, driving @ Usual, pre-accident Speed R, he covers 90 km in (3/2) hours
R = (90 km) / (3/2 hours) = 60 km/hr
which means
(3/4)R = 45 km/hr = Reduced Speed after the Accident
(2nd) Using the Regular Speed R = 60 and the Reduced Speed = 45 ------> find the Total Distance of the Trip
First, just focus on the portion AFTER the Accident took place in Scenario 1 above.
in Scenario 1, call the Distance from the Accident {A} to the Destination = D
had he NOT got into an accident, he would have covered this D driving @ 60 km/hr without being Late. Call the Usual "on-time" Time = T
because he actually DID get into this accident, he covered this D driving
a Reduced Speed of 45 km/hr and was + 3 hours Late (Removing the 1/2 hour lay-over which does NOT Contribute to the Distance)
Time taken
45 km/hr = (Time @ 60 km/hr) + 3 more hours of "lateness"
Time = Distance/Speed
D/45 = D/60 + 3 ------------- where D is in km and the Time is in hours
(D/45) - (D/60) = 3
(60D - 45D) / (2,700) = 3
(15) / (2,700) * D = 3
D = (3 * 2,700) / (15)
D = 540 km = the Distance form the point of the Accident to the Final Destination
There was ALSO 1 hour that he drove before he had the accident
he drove for 1 hour at his Usual, pre-accident Speed = 60 km/hr
Total Distance = 600 km
Finally, need to convert to miles-------> 1 km = .6213 mile
we can approximate:
(600 km) * (.62) = approximately 372 miles (which is estimated a LITTLE Under the Actual Value)
-B-
373 Miles