SOLUTION: a train an a car leave at the same time 8:00 am on a 156 mile trip the train travels at 80 mph with a 30 minute layover the car travels at 55 mph which will arrive first." please s
Question 387045: a train an a car leave at the same time 8:00 am on a 156 mile trip the train travels at 80 mph with a 30 minute layover the car travels at 55 mph which will arrive first." please show equation " Found 2 solutions by ewatrrr, CharlesG2:Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
Hi,
D = r*t Or D/r = t
t(car) = 156mi/55mph = 2.84 hr
T(train) = = 156mi/80mph + .5hr = 1.95 + .5 = 2.45 hr
Even with the layover, the train will arrive first.
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website! "a train an a car leave at the same time 8:00 am on a 156 mile trip the train travels at 80 mph with a 30 minute layover the car travels at 55 mph which will arrive first." please show equation ""
Distance = Rate * Time
Car's Distance = Train's Distance = 156 miles
Car's Rate = 55 mph = CR
Train's Rate = 80 mph = TR
30 minutes = 0.5 hours
Car's Time --> 156 miles / 55 mph = 2.8363 (63 repeating) hours
156 - (55 * 2) = 156 - 110 = 46 --> 46/55 of 60 minutes -->
50.1818 (18 repeating) minutes --> car takes 2 hours and 50 minutes -->
so car arrives 10:50 AM
Train's Time Nonstop --> 156 miles / 80 mph = TTN = 1.95
Train's Time = TT = 1.95 + 0.5 = 2.45 hours -->
0.45 * 60 minutes = 27 minutes -->
train takes 2 hours and 27 minutes --> train arrives at 10:27 AM