My friend sets out walking at 3 miles per hr. I set out behind her 5 minutes later at 4 miles per hour. For how many minutes will my friend have been walking when I catch up with her?

- anonymous

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- anonymous

I think that you need to do 3*1/12=1/4 mile which is the distance between them after 5 minutes.

- anonymous

then time it takes for them to catch up by 1/4 mile is 12

- anonymous

is it not?

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## More answers

- anonymous

because every hour there is a disparity of 1 mile between them.

- anonymous

so for them to cover 1/4 mile they would need 1/4 hour .

- anonymous

60*1/4=15

- anonymous

yeah sorry about that it's I think 15min your thoughts?

- kropot72

Let d be the distance walked before catch up.
Time in minutes taken by friend at catch up = d/3 * 60
Time in minutes taken by me at catch up = d/4 + 5
Therefore equating the times we get
\[\large \frac{d}{3}\times60=\frac{d}{4}\times60+5\]
Now if you solve to find the value of d, you can plug the d value into d/3 * 60 to find the time to catch up in minutes.

- anonymous

Since the slower guy with 3 mile/per hour walked for 5 minutes, it's 3*1/12=1/4

- kropot72

d turns out to be 1 mile. Therefore the required time is 20 minutes.

- kropot72

@danyboy9169 Do you understand?

- anonymous

I don't think so

- anonymous

By the time 4 mile guy starts off there is a 1/4 mile distance

- kropot72

What part don't you understand?

- anonymous

because 5/12=1/12
1/12*3miles/hour=1/4 mile between them

- UsukiDoll

we're using distance formula aren't we ?

- anonymous

It's physics

- anonymous

1/4mile to be covered with 1mile/h is of course it takes 1/4h

- anonymous

60*1/4=15min

- anonymous

Or if you taken the extra 5 min from the friend it will be 20min

- anonymous

is it not?

- kropot72

"Time in minutes taken by me at catch up = d/4 * 60 + 5"
Correction to my first post. The rest is correct.

- UsukiDoll

might as well create a table
Friend - 3 mph (speed)
me - 4mph (speed) 5 minutes later so there's a delay

- anonymous

It's classical kinematics physics question

- UsukiDoll

1 hour = 60 minutes.

- anonymous

You convert 5 min to hour.
5min*1hour/60min=

- UsukiDoll

so we're solving for the distance

- anonymous

units "min" cancel out, leaving out the 1/12h.

- UsukiDoll

mhm always remember those labels XD

- anonymous

Everything makes sense with units.

- anonymous

3mile/hour*1/12h=3/12mile(since h cancels out) or 1/4mile

- anonymous

There is 1/4mile between the friends

- UsukiDoll

\[\large \frac{d}{3}\times60=\frac{d}{4}\times60+5\]
\[\frac{60d}{3}=\frac{60d}{4}+5\]
\[20d=15d+5\]
\[20d-15d=5\]
\[5d=5\]
\[d=1\]

- anonymous

4mile/hour-3mile/hour=difference of 1mile/hour

- anonymous

for 1 mile/hour to catch up for 1/4 mile

- kropot72

Let d be the distance walked before catch up.
Time in minutes taken by friend at catch up = d/3 * 60
Time in minutes taken by me at catch up = d/4 * 60 + 5 Therefore equating the times we get
\[\large \frac{d}{3}\times60=\frac{d}{4}\times60+5\]
which simplifies to
\[\large 20d=15d+5\]
therefore d = 1 mile
The time taken for the friend to walk 1 mile at 3 mph = 20 minutes.

- anonymous

d=1/4(miles)
v=1mile/h

- UsukiDoll

so going back a bit if d = 1 then both sides would be 20
1 mile
20 minutes

- anonymous

Use the kinematics formula t=d
t=0.25mile/1mile/h=1/4h=15 minutes. Plus that 5 min for initial dilation.
15min+5min=20min

- anonymous

Use physics formula it's much easier

- kropot72

@Robert136
"Use the kinematics formula t=d"
A new one on me!!!!!!

- UsukiDoll

:P nah prefer da mathz

- kropot72

How about
\[\large time=\frac{distance}{speed}\]

- anonymous

Yup

- anonymous

You can manipulate the formula without thinking much. Only the algebra skills required

- UsukiDoll

yesssssssssssssssss @kropot72

- anonymous

So any missing value will be compensated by rearranging the formula

- anonymous

Got it, Since I walk 1 mile per hour faster than my friend, the distance between us shrinks at a rate of 1 mile per hour.
Since I am 1/4 mile behind when I start walking, it takes me 1/4 hour (= 15 minutes) to close the 1/4-mile gap.
At that time, my friend, who started 5 minutes before me, will have been walking for 5+15 =20 minutes.

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