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Sam left his school at 3:05. He walked at a speed of 3.2 mph. 15 minutes later, Al started running after him, and he caught up with Sam 10 minutes later. What was Al’s speed?

joebob  Jan 21, 2018
 #1
avatar+14579 
+2

Sam has a head start of   3.2 m/hr  x 1/4 hr = .8 miles

 

Al has to cover this .8 miles PLUS the distance Sam covers in an additional 10 minutes

Sam's additional distance in 10 minutes   =  3.2 m/hr x 10/60 hr = .5333 mile

 

So AL has to cover   .8 mile + .5333 miles in 10 minutes to catch Sam

 

(.8mile + .5333 mile) / (10/60  hr)   = 8 m/hr

ElectricPavlov  Jan 21, 2018
 #2
avatar
+2

Distance = Rate(speed) x Time

Sam' s time = 15 + 10 = 25 minutes

Distance covered by Sam:

3.2 mph x 25/60 =1 1/3 miles

 

Let Al's speed=S

Distance covered by Al = S mph x 10/60 =1/6S miles

1 1/3 miles = 1/6S miles, solve for S

S =[1 1/3] / 1/6

S=8 mph - Al's speed.

Guest Jan 21, 2018

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