Click here to go to the

By Matlm (Matlm) on Monday, October 20, 2003 - 08:33 pm: Edit |

you know, the typical Sally traveled 30 mph goign to the mall, and 60 mph coming home. what was her average speed.

By Curiousone (Curiousone) on Monday, October 20, 2003 - 08:35 pm: Edit |

for that if its the same distance both ways isn't it just (30 + 60)/2?

Otherwise if it gives you specific distances you have to find the total distance overall and divide it by the total amount of time.

By Justice (Justice) on Monday, October 20, 2003 - 09:00 pm: Edit |

No it's not. Remember, the key is distance = rate x time. You're looking for rate when you find average speed, so find the total distance traveled, which should be easy, and the total time traveled. You shouldn't be doing any division over two to find average speed...

By Matlm (Matlm) on Monday, October 20, 2003 - 09:03 pm: Edit |

how do you find total distance?

By Montydsw11 (Montydsw11) on Monday, October 20, 2003 - 09:05 pm: Edit |

2ab/(a+b)....

3600/90

=40mph

Thank you, you can hold your applause...

lol jk

By Montydsw11 (Montydsw11) on Monday, October 20, 2003 - 09:09 pm: Edit |

matlm you can solve this problem a number of ways. If you feel more comfortable doing it the "slow" way as described by the folks above, I can show you how to do that also, but my way is a lot better.

Let me know if you ever need any sort of math help for the SAT, feel free to send me an email or something... (that goes for everyone else on this board)

By Curiousone (Curiousone) on Monday, October 20, 2003 - 09:42 pm: Edit |

**No it's not. Remember, the key is distance = rate x time. You're looking for rate when you find average speed, so find the total distance traveled, which should be easy, and the total time traveled. You shouldn't be doing any division over two to find average speed... **

Which is exactly what I said on the second line. IF IT'S THE SAME DISTANCE BOTH WAYS, yes, you can add the rates up and divide in half because there's a constant.

By Sidis (Sidis) on Monday, October 20, 2003 - 09:45 pm: Edit |

Hey those questions are always annoying! Please show us the quick way! thanks.

By Justice (Justice) on Monday, October 20, 2003 - 09:52 pm: Edit |

No curiousone you're completely incorrect. Ok let's say that a dude goes to a city and drives back. The distance from where he starts to the city is 100 miles. He goes there traveling at 1mph and returns at 100mph. Certainly his average speed is not 50.5mph...you should know that from common sense. Whether or not the distance is constant does not matter because the time is not constant; therefore you should never be dividing by 2, like I said.

The real average speed would be 200 = (1+100)(rate)

rate = 1.98mph.

By Sidis (Sidis) on Monday, October 20, 2003 - 10:05 pm: Edit |

Ok, to set it straigh solve this:

Jhonny goes from his house(a) to house(b), 50mph. Then he goes back at 70mph. The distance between A and B is 60 miles.

What is the total time?

What is the average speed?

Thanks

By Matlm (Matlm) on Monday, October 20, 2003 - 10:10 pm: Edit |

Hey monty, is your formula from physics???

By Perry2006 (Perry2006) on Monday, October 20, 2003 - 10:36 pm: Edit |

Hehe, I noticed that formula when I was doing AoPS as well.

a = speed going to an area

b = speed returning

d = distance

Total distance / total time = AVE. speed

2d/(d/a + d/b) = 2d/[d(a+b)/ab] = 2ab/(a+b)

By Justice (Justice) on Monday, October 20, 2003 - 11:17 pm: Edit |

60/50 + 60/70 = total time = 2.057 hrs

total distance = 2(60) = 120 mi

120mi/2.057hrs = 58.3mph

By Sidis (Sidis) on Monday, October 20, 2003 - 11:43 pm: Edit |

well, that was better than the creepier explanations. I get it now thanks

By Montydsw11 (Montydsw11) on Tuesday, October 21, 2003 - 12:01 am: Edit |

matlm... my formula is from common sense. Unfortunittly that is hard to come by on this board...

Posting is currently disabled in this topic. Contact your discussion moderator for more information. |

Administrator's Control Panel -- Board Moderators Only Administer Page |