SOLUTION: Hi John and Kevin travel from home to school using the same route. They started at the same time. John travelled at 18km per hour. Both did not change their speeds throughout the

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Question 1201263: Hi
John and Kevin travel from home to school using the same route. They started at the same time. John travelled at 18km per hour. Both did not change their speeds throughout the trip. When Kevin covered 1/4 of the journey, John was 3.5km ahead of him. If John reached school at 9.35am what time did Kevin reach school.
Thanks

Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
John and Kevin travel from home to school using the same route.
They started at the same time. John travelled at 18km per hour.
Both did not change their speeds throughout the trip.
When Kevin covered 1/4 of the journey, John was 3.5km ahead of him.
If John reached school at 9.35am what time did Kevin reach school.
~~~~~~~~~~~~~~~~~~

Let x be the distance from home to the school, in kilometers;

Let J be the John' speed, in km/h;  let K be the Kevin' speed.

Let' start our clock when both John and Kevin left their home.



    From the condition, we can write this equation

        %28x%2F4%2B+3.5%29%2FJ = %28%28x%2F4%29%29%2FK,

    saying that "When Kevin covered 1/4 of the journey, John was 3.5km ahead of him."



We know that J = 18 km/h, so we replace J in this equation by 18. We get then

    %28x%2F4%2B3.5%29%2F18 = %28%28x%2F4%29%29%2FK.


Multiply both sides by 4. You will get

    %28x+%2B+14%29%2F18 = x%2FK.


The time x%2FK is the time of Kevin's travel and is the time under the problem's question.

The time %28x%2B14%29%2F18  is  14%2F18 = 7%2F9 of an hour more than x%2F18,
which is the time when John arrived to the school, i.e. 9:35 am.


7%2F9  of an hour is %287%2A3600%29%2F9 seconds = 7*400 seconds = 2800 seconds = 46 minutes and 40 seconds.


Hence, Kevin reaches the school 46 minutes and 40 seconds after 9:35 am.


ANSWER.  Kevin reaches the school at 10:21 minutes and 40 seconds A.M.

Solved.


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It can be re-told in other words.

One can say that under the given condition  

    "when Kevin covered 1/4 of the journey, John was 3.5km ahead of him"

we have this logical implication

    "when Kevin covers the entire journey, John is 4*3.5 km = 14 kilometers ahead of him."


Hence, Kevin reaches the school in that additional time after 9:35 am,
when John covers 14 kilometers, moving at the rate of 18 km/h.

It leads you to the same answer.



Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Answer is 10:21:40 AM

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.

=========================================================================

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 14 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 = 14

The key single number to pull out of all this is the 14.
It will come in handy in the next section below.

------------------------------------------

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.


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.

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:


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.

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 10:21:40 AM
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.