anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I think that you need to do 3*1/12=1/4 mile which is the distance between them after 5 minutes.
anonymous
  • anonymous
then time it takes for them to catch up by 1/4 mile is 12
anonymous
  • anonymous
is it not?

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anonymous
  • anonymous
because every hour there is a disparity of 1 mile between them.
anonymous
  • anonymous
so for them to cover 1/4 mile they would need 1/4 hour .
anonymous
  • anonymous
60*1/4=15
anonymous
  • anonymous
yeah sorry about that it's I think 15min your thoughts?
kropot72
  • 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
  • anonymous
Since the slower guy with 3 mile/per hour walked for 5 minutes, it's 3*1/12=1/4
kropot72
  • kropot72
d turns out to be 1 mile. Therefore the required time is 20 minutes.
kropot72
  • kropot72
@danyboy9169 Do you understand?
anonymous
  • anonymous
I don't think so
anonymous
  • anonymous
By the time 4 mile guy starts off there is a 1/4 mile distance
kropot72
  • kropot72
What part don't you understand?
anonymous
  • anonymous
because 5/12=1/12 1/12*3miles/hour=1/4 mile between them
UsukiDoll
  • UsukiDoll
we're using distance formula aren't we ?
anonymous
  • anonymous
It's physics
anonymous
  • anonymous
1/4mile to be covered with 1mile/h is of course it takes 1/4h
anonymous
  • anonymous
60*1/4=15min
anonymous
  • anonymous
Or if you taken the extra 5 min from the friend it will be 20min
anonymous
  • anonymous
is it not?
kropot72
  • kropot72
"Time in minutes taken by me at catch up = d/4 * 60 + 5" Correction to my first post. The rest is correct.
UsukiDoll
  • UsukiDoll
might as well create a table Friend - 3 mph (speed) me - 4mph (speed) 5 minutes later so there's a delay
anonymous
  • anonymous
It's classical kinematics physics question
UsukiDoll
  • UsukiDoll
1 hour = 60 minutes.
anonymous
  • anonymous
You convert 5 min to hour. 5min*1hour/60min=
UsukiDoll
  • UsukiDoll
so we're solving for the distance
anonymous
  • anonymous
units "min" cancel out, leaving out the 1/12h.
UsukiDoll
  • UsukiDoll
mhm always remember those labels XD
anonymous
  • anonymous
Everything makes sense with units.
anonymous
  • anonymous
3mile/hour*1/12h=3/12mile(since h cancels out) or 1/4mile
anonymous
  • anonymous
There is 1/4mile between the friends
UsukiDoll
  • 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
  • anonymous
4mile/hour-3mile/hour=difference of 1mile/hour
anonymous
  • anonymous
for 1 mile/hour to catch up for 1/4 mile
kropot72
  • 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
  • anonymous
d=1/4(miles) v=1mile/h
UsukiDoll
  • UsukiDoll
so going back a bit if d = 1 then both sides would be 20 1 mile 20 minutes
anonymous
  • 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
  • anonymous
Use physics formula it's much easier
kropot72
  • kropot72
@Robert136 "Use the kinematics formula t=d" A new one on me!!!!!!
UsukiDoll
  • UsukiDoll
:P nah prefer da mathz
kropot72
  • kropot72
How about \[\large time=\frac{distance}{speed}\]
anonymous
  • anonymous
Yup
anonymous
  • anonymous
You can manipulate the formula without thinking much. Only the algebra skills required
UsukiDoll
  • UsukiDoll
yesssssssssssssssss @kropot72
anonymous
  • anonymous
So any missing value will be compensated by rearranging the formula
anonymous
  • 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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