SOLUTION: Martina leaves home at 9 A.M., bicycling at a rate of
24 mi/h. Two hours later, John leaves, driving at the rate of 48 mi/h. At what time
will John catch up with Martina?
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-> SOLUTION: Martina leaves home at 9 A.M., bicycling at a rate of
24 mi/h. Two hours later, John leaves, driving at the rate of 48 mi/h. At what time
will John catch up with Martina?
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Question 106148This question is from textbook beginning algebra
: Martina leaves home at 9 A.M., bicycling at a rate of
24 mi/h. Two hours later, John leaves, driving at the rate of 48 mi/h. At what time
will John catch up with Martina? This question is from textbook beginning algebra
You can put this solution on YOUR website!
distance(d)=rate(r) times time(t) or d=rt;t=d/r and r=d/t
Let t=time that elapses after John leaves until he catches Martina
We know that when John leaves, Martina has already travelled 24*2 or 48 miles (d=rt)
Distance Martina has travelled when John catches her=48+24t
Distance John has travelled when he catches Martina=48t
Now we know that John will have caught up with Martina when they have both travelled the same distance.
So our equation to solve is:
48+24t=48t subtract 24t from both sides
48+24t-24t=48t-24t collect like terms
48=24t divide both sides by 24
t=2 hrs---------------------time that elapses after John leaves until he catches Martina
We are also told that John leaves at 11:00AM.
So 11:00AM plus 2 hours=1:00PM ---------time that John catches Martina
CK
48+24*2=48*2
48+48=96
96=96
Hope this helps---ptaylor
