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Liana's class was going on a very exciting school trip, but when Liana got to school, the school bus had already left! Frantic, Liana begged her mom to drive her to catch up with the bus. Liana's mother has to decide quickly whether she can do it. These are the facts to consider: The bus left at 8:00 AM and is moving with an average speed of 45 mph. Liana and her mom can leave no earlier than at 8:40. If Liana's mother drives with the average speed 60 mph, when can they catch up with the bus? 

 Jan 27, 2020
 #1
avatar+33615 
+4

Let t be the number of hours since 8.00.  The bus travels  45t miles in that time.  Liana travels 60*(t - 2/3) miles in that time (noting that 40 minutes is 2/3 of an hour).  Hence we want 45t = 60(t - 2/3), so t = 8/3 hours, or 2 hours 40 minutes.  They catch up at 10.40.

 Jan 27, 2020
 #2
avatar+36916 
+2

The bus gets a 40 min   ( 2/3 hour)   head start    2/3 hr x 45 m/hr = 30 mile head start

Mom drives  (60-45) = 15 m/hr faster than the bus....it will take 2 hours to close the 30 mile headstart gap.

8:40  + 2 hours  = 10 :40

 Jan 27, 2020

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