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anonymous
 one year ago
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
 one year ago
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?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think that you need to do 3*1/12=1/4 mile which is the distance between them after 5 minutes.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then time it takes for them to catch up by 1/4 mile is 12

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because every hour there is a disparity of 1 mile between them.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for them to cover 1/4 mile they would need 1/4 hour .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah sorry about that it's I think 15min your thoughts?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1Let 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
 one year ago
Best ResponseYou've already chosen the best response.0Since the slower guy with 3 mile/per hour walked for 5 minutes, it's 3*1/12=1/4

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1d turns out to be 1 mile. Therefore the required time is 20 minutes.

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1@danyboy9169 Do you understand?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0By the time 4 mile guy starts off there is a 1/4 mile distance

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1What part don't you understand?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because 5/12=1/12 1/12*3miles/hour=1/4 mile between them

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1we're using distance formula aren't we ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01/4mile to be covered with 1mile/h is of course it takes 1/4h

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Or if you taken the extra 5 min from the friend it will be 20min

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1"Time in minutes taken by me at catch up = d/4 * 60 + 5" Correction to my first post. The rest is correct.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1might as well create a table Friend  3 mph (speed) me  4mph (speed) 5 minutes later so there's a delay

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's classical kinematics physics question

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.11 hour = 60 minutes.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You convert 5 min to hour. 5min*1hour/60min=

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1so we're solving for the distance

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0units "min" cancel out, leaving out the 1/12h.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1mhm always remember those labels XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Everything makes sense with units.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03mile/hour*1/12h=3/12mile(since h cancels out) or 1/4mile

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There is 1/4mile between the friends

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1\[\large \frac{d}{3}\times60=\frac{d}{4}\times60+5\] \[\frac{60d}{3}=\frac{60d}{4}+5\] \[20d=15d+5\] \[20d15d=5\] \[5d=5\] \[d=1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.04mile/hour3mile/hour=difference of 1mile/hour

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for 1 mile/hour to catch up for 1/4 mile

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1Let 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
 one year ago
Best ResponseYou've already chosen the best response.0d=1/4(miles) v=1mile/h

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1so going back a bit if d = 1 then both sides would be 20 1 mile 20 minutes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Use 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
 one year ago
Best ResponseYou've already chosen the best response.0Use physics formula it's much easier

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1@Robert136 "Use the kinematics formula t=d" A new one on me!!!!!!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1:P nah prefer da mathz

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1How about \[\large time=\frac{distance}{speed}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can manipulate the formula without thinking much. Only the algebra skills required

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1yesssssssssssssssss @kropot72

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So any missing value will be compensated by rearranging the formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Got 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 1/4mile 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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