A person takes a trip, driving with a constant speed of 90.5 km/h except for a 21.0 min rest stop. If the person's average speed is 63.1 km/h, how much time is spent on the trip and how far does the person travel?
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We start with distance = average speed X time to cover that distance; that's S = Vavg T.
Vavg = 63.1 kph. Total trip time = T = dwell time Td + travel time Tt = 21/60 hrs + S/90.5.
Put this all together and we have S = Vavg (21/60 + S/90.5) = 21*63.1/60 + 63.1 S/90.5. Then
S(1 - 63.1/90.5) = 21*63.1/60 ans, ta da, S = 21*63.1/((1 - 63.1/90.5)*60) = 72.94498175 = 72.9 km. ANS. NOTE the extra 60 to convert 21 mins to hrs.
Then the total time T = S/Vavg = 72.9/63.1 = 1.155309033 = 1.15 hrs. ANS.
Vavg = 63.1 kph. Total trip time = T = dwell time Td + travel time Tt = 21/60 hrs + S/90.5.
Put this all together and we have S = Vavg (21/60 + S/90.5) = 21*63.1/60 + 63.1 S/90.5. Then
S(1 - 63.1/90.5) = 21*63.1/60 ans, ta da, S = 21*63.1/((1 - 63.1/90.5)*60) = 72.94498175 = 72.9 km. ANS. NOTE the extra 60 to convert 21 mins to hrs.
Then the total time T = S/Vavg = 72.9/63.1 = 1.155309033 = 1.15 hrs. ANS.