Question 1011578: John can walk at 5 km/h and run at 15 km/h. A and B are two stops on a road. Both are 2 km from John’s home H. Draw a diagram.
a) The bus leaves from A in 15 minutes. How far can John walk from home before he has to run to catch the bus?
b) It takes the bus 3 minutes to travel from A to B. If John catches the bus at B, how much less running does John have to do?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website!
in order to solve this, you need to either convert everything to minutes or convert everything to hours.
we'll convert everything to hours.
15 minutes is equal to .25 hours.
there are two pieces to the journey.
there's the piece that he walks and the piece that he rides.
we'll call the piece that he walks d1.
we'll call the piece that he runs d2.
we'll call t1 the time that it takes to walk d1 kilometers.
we'll call t2 the time that it takes to walk d2 kilometers.
to get to point A, the total time required is 15 minutes which is equal to .25 hours.
since rate * time = distance, we have the following formulas to get to point A.
t1 + t2 = .25 hours
d1 + d2 = 2 kilometers
d1 = r1*t1
d2 = r2*t2
r1 = 5 kilometers per hour.
r2 = 15 kilometers per hour.
we get:
d1 = r1*t1 becomes d1 = 5*t1
d2 = r2*t2 becomes d2 = 15*t2
d1 + d2 = 2 becomes 5*t1 + 15*t2 = 2
since t1 + t2 = .25, we can solve for t2 to get t2 = .25 - t1.
in the equation of 5*t1 + 15*t2 = 2, we replace t2 with .25 - t1 to get:
5*t1 + 15*(.25-t1) = 2
simplify to get 5*t1 + 15*.25 - 15*t1 = 2
simplify further to get 5*t1 + 3.75 - 15*t1 = 2
combine like terms to get -10*t1 + 3.75 = 2
subtract 3.75 from both sides of the equation to get -10*t1 = -1.75
divide both sides of the equation by -10 to get t1 = -1.75/-10 = .175
we have t1 = .175
since t2 = .25 - t1, we get t2 = .25 - .175 = .075
since d1 = r1*t1, we get d1 = 5*.175 = .875
since d2 = r2*t2, we get d2 = 15*.075 = 1.125
in order to get to point A in 15 minutes, he can walk .875 kilometers and then has to run 1.125 kilometers for a total distance of 2 kilometers.
when he is going to B, he has another 3 minutes to play with.
he needs to get to B in 18 minutes rather than 15 minutes.
this is because the bus leaves from A in 15 minutes and takes another 3 minutes to get to B.
18 minutes is equal to 18/60 = .3 hours.
you will do the same calculations, except now you will have:
t1 + t2 = .3 rather than t1 + t2 = .25
you can solve for t2 to get t2 = .3 - t1
your calculations will now become:
since t1 + t2 = .3, we can solve for t2 to get t2 = .3 - t1.
in the equation of 5*t1 + 15*t2 = 2, we replace t2 with .3 - t1 to get:
5*t1 + 15*(.3-t1) = 2
simplify to get 5*t1 + 15*.3 - 15*t1 = 2
simplify further to get 5*t1 + 4.5 - 15*t1 = 2
combine like terms to get -10*t1 + 4.5 = 2
subtract 4.5 from both sides of the equation to get -10*t1 = -2.5
divide both sides of the equation by -10 to get t1 = -2.5/-10 = .25
we have t1 = .25
since t2 = .3 - t1, we get t2 = .3 - .25 = .05
since d1 = r1*t1, we get d1 = 5*.25 = 1.25
since d2 = r2*t2, we get d2 = 15*.05 = .75
in order to get to point B in 18 minutes, he can walk 1.25 kilometers and then has to run .75 kilometers for a total distance of 2 kilometers.
if john catches the bus at A, he has to run 1.125 kilometers.
if john catches the bus at B, he has to run .75 kilometers.
if john catches the bus at B rather than A, he has to run 1.125 - .75 = .375 kilometers less.
since you know that he needs to get to point A in 15 minutes and he needs to get to point B in 18 minutes and that he walks 5 kilometers per hour and that he runs 15 kilometers per hour, you can confirm you solution is correct by doing the following.
since rate * time = distance, then time = distance / rate.
going to point A, you get .875/5 + 1.125/15 = .25 hours * 60 = 15 minutes.
.875 + 1.125 = 2 kilometers.
going to point B, you get 1.25/5 + .75/15 = .3 hours * 60 = 18 minutes.
1.25 + .75 = 2 kilometers.
numbers check out so solution looks good.
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