Question 1201263
<font color=black size=3>
Answer is <font color=red size=4>10:21:40 AM</font>


This is in the format of hours:minutes:seconds
If the seconds don't really matter then we could round up to the nearest minute to get 10:22 AM.
The timestamp of 10:21 AM wouldn't work because Kevin would be slightly short of arriving at his goal.


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Explanation:



Let's think of Kevin's journey in terms of a progress bar. 
I'll use percentages for the progress bar.


0% means Kevin is at home.
100% means Kevin made it to school.


Attached to each percentage, I'll indicate how far John is ahead of Kevin.


0% -- John is 0 km ahead
25% -- John is 3.5 km ahead
50% -- John is 7 km ahead
75% -- John is 10.5 km ahead
100% -- John is <font color=red>14</font> km ahead


Scratch Work:
3.5*0 = 0
3.5*1 = 3.5
3.5*2 = 7
3.5*3 = 10.5
3.5*4 = <font color=red>14</font>


The key single number to pull out of all this is the <font color=red>14</font>. 
It will come in handy in the next section below.


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Now let's draw out what's going on.
The drawing is optional, but it might be helpful for visual learners.


First we define the following locations
A = home's location
B = 25% of the distance from home to school
C = Kevin's location 3.5 km ahead of point B
D = school's location
E = I'll explain this point later


Here's a snapshot of the start of the journey.
Both men start at location A on the number line.
{{{
drawing(400,300,1,9,2,8,
graph(400,300,1,9,2,8),
locate(1,7.5,matrix(1,4,"Kevin's","progress","=","0%")),
line(-3,5,13,5),line(2,5.2,2,4.8),line(3.2,5.2,3.2,4.8),line(4.4,5.2,4.4,4.8),line(5.6,5.2,5.6,4.8),line(8,5.2,8,4.8),locate(1.95,4.75,"A"),locate(3.15,4.75,"B"),locate(4.35,4.75,"C"),locate(5.55,4.75,"D"),locate(7.95,4.75,"E"),
red(
line(2,3.5,2,4.3),line(2,4.3,2.08,4.1614),line(2,4.3,1.92,4.1614),
locate(1.75,3.5,John)
),
blue(
line(2,6.7,2,5.5),line(2,5.5,1.88,5.7078),line(2,5.5,2.12,5.7078),
locate(1.75,7,Kevin)
),
locate(1,2.7,matrix(1,2,Diagram,not)),
locate(1,2.35,matrix(1,2,to,scale))
)
}}}


Let's fast-forward to where Kevin travels 1/4 = 25% of the journey.
Kevin will move to point B.
John moves to point C, which is 3.5 km further ahead compared to B.
{{{
drawing(400,300,1,9,2,8,
locate(1,7.5,matrix(1,4,"Kevin's","progress","=","25%")),
line(-3,5,13,5),line(2,5.2,2,4.8),line(3.2,5.2,3.2,4.8),line(4.4,5.2,4.4,4.8),line(5.6,5.2,5.6,4.8),line(8,5.2,8,4.8),locate(1.95,4.75,"A"),locate(3.15,4.75,"B"),locate(4.35,4.75,"C"),locate(5.55,4.75,"D"),locate(7.95,4.75,"E"),
red(
line(4.4,3.5,4.4,4.3),line(4.4,4.3,4.48,4.1614),line(4.4,4.3,4.32,4.1614),locate(4.15,3.5,John)
),
blue(
line(3.2,6.7,3.2,5.5),line(3.2,5.5,3.08,5.7078),line(3.2,5.5,3.32,5.7078),locate(2.95,7,Kevin)
),
locate(1,2.7,matrix(1,2,Diagram,not)),
locate(1,2.35,matrix(1,2,to,scale))
)
}}}
The gap from B to C is 3.5 km.
BC = 3.5



Now let's fast-forward to when John arrives at school.
John moves to point D and it is 9:35 AM.
Kevin is somewhere between B and D, excluding either endpoint.
This is what that looks like:
{{{
drawing(400,300,1,9,2,8,
locate(1,7.5,matrix(1,8,"Kevin's","progress","=","somewhere","between","25%","and","100%")),
locate(5,7,matrix(1,2,"(excluding","endpoints)")),
line(-3,5,13,5),line(2,5.2,2,4.8),line(3.2,5.2,3.2,4.8),line(4.4,5.2,4.4,4.8),line(5.6,5.2,5.6,4.8),line(8,5.2,8,4.8),locate(1.95,4.75,"A"),locate(3.15,4.75,"B"),locate(4.35,4.75,"C"),locate(5.55,4.75,"D"),locate(7.95,4.75,"E"),
red(
line(5.6,3.5,5.6,4.3),line(5.6,4.3,5.68,4.1614),line(5.6,4.3,5.52,4.1614),locate(5.35,3.5,John),locate(5,2.35,matrix(1,2,"9:35","AM"))
),
blue(
arc(3.44,5.44,0.48,0.48,180,270),line(3.44,5.68,4.16,5.68),arc(4.16,5.92,0.48,0.48,0,90),arc(4.64,5.92,0.48,0.48,90,180),line(4.64,5.68,5.36,5.68),arc(5.36,5.44,0.48,0.48,270,360),
locate(4.15,6.5,Kevin)
),
locate(1,2.7,matrix(1,2,Diagram,not)),
locate(1,2.35,matrix(1,2,to,scale))
)
}}}


Technically John should stop when reaching point D.
However, I'll have him keep moving to reach point E.


Once John reaches point E, Kevin will have arrived at point D.
{{{
drawing(400,300,1,9,2,8,
locate(1,7.5,matrix(1,4,"Kevin's","progress","=","100%")),
line(-3,5,13,5),line(2,5.2,2,4.8),line(3.2,5.2,3.2,4.8),line(4.4,5.2,4.4,4.8),line(5.6,5.2,5.6,4.8),line(8,5.2,8,4.8),locate(1.95,4.75,"A"),locate(3.15,4.75,"B"),locate(4.35,4.75,"C"),locate(5.55,4.75,"D"),locate(7.95,4.75,"E"),
red(
line(8,3.5,8,4.3),line(8,4.3,8.08,4.1614),line(8,4.3,7.92,4.1614),locate(7.75,3.5,John),
locate(5,2.35,matrix(1,5,some,time,after,"9:35","AM"))
),
blue(
line(5.6,6.7,5.6,5.5),line(5.6,5.5,5.48,5.7078),line(5.6,5.5,5.72,5.7078),locate(5.35,7,Kevin)
),
locate(6.5,5.4, matrix(1,2,14,km)),
locate(1,2.7,matrix(1,2,Diagram,not)),
locate(1,2.35,matrix(1,2,to,scale))
)
}}}
Point E is 14 km ahead of point D.
This was the 14 I mentioned earlier.


Let's determine how long it takes for John to travel that extra 14 km.
distance = rate*time
time = distance/rate
time = 14/18
time = 7/9 of an hour
This is the time it takes for John to move from point D to point E.
This time value is added onto the time of 9:35 AM to determine when Kevin arrives at school.


To have 7/9 of an hour make more sense, let's first convert to minutes:seconds format.
1 hr = 60*60 = 3600 sec
7/9 hrs = (7/9)*3600 = 2800 sec
2800/60 = 46.6666666666667
46 min = 46*60 = 2760 sec
2800 sec = 2760 sec + 40 sec
2800 sec = 46 min + 40 sec
Therefore,
7/9 of an hour = 46 min + 40 sec


We'll add the duration "46 min + 40 sec" onto the starting time of 9:35 AM to determine the final answer.


The jump from 9:35 AM to 10:00 AM is 25 minutes.
The remaining 46-25 = 21 minutes has us go from 10:00 AM to 10:21 AM
Then we have an additional 40 seconds tacked on at the end.



Final Answer is <font color=red size=4>10:21:40 AM</font>
This is in the format of hours:minutes:seconds
If the seconds don't really matter then we could round up to the nearest minute to get 10:22 AM.
The timestamp of 10:21 AM wouldn't work because Kevin would be slightly short of arriving at his goal.
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