## 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?

1. anonymous

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

2. anonymous

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

3. anonymous

is it not?

4. anonymous

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

5. anonymous

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

6. anonymous

60*1/4=15

7. anonymous

8. 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.

9. anonymous

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

10. kropot72

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

11. kropot72

@danyboy9169 Do you understand?

12. anonymous

I don't think so

13. anonymous

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

14. kropot72

What part don't you understand?

15. anonymous

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

16. UsukiDoll

we're using distance formula aren't we ?

17. anonymous

It's physics

18. anonymous

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

19. anonymous

60*1/4=15min

20. anonymous

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

21. anonymous

is it not?

22. kropot72

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

23. UsukiDoll

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

24. anonymous

It's classical kinematics physics question

25. UsukiDoll

1 hour = 60 minutes.

26. anonymous

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

27. UsukiDoll

so we're solving for the distance

28. anonymous

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

29. UsukiDoll

mhm always remember those labels XD

30. anonymous

Everything makes sense with units.

31. anonymous

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

32. anonymous

There is 1/4mile between the friends

33. UsukiDoll

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

34. anonymous

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

35. anonymous

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

36. 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.

37. anonymous

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

38. UsukiDoll

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

39. 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

40. anonymous

Use physics formula it's much easier

41. kropot72

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

42. UsukiDoll

:P nah prefer da mathz

43. kropot72

How about $\large time=\frac{distance}{speed}$

44. anonymous

Yup

45. anonymous

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

46. UsukiDoll

yesssssssssssssssss @kropot72

47. anonymous

So any missing value will be compensated by rearranging the formula

48. 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.