Ask Me Help Desk

Ask Me Help Desk (https://www.askmehelpdesk.com/forum.php)
-   Mathematics (https://www.askmehelpdesk.com/forumdisplay.php?f=199)
-   -   Speed Equation (https://www.askmehelpdesk.com/showthread.php?t=64926)

  • Feb 21, 2007, 09:24 AM
    dylanjacobus
    Speed Equation
    Hey, I am having a hard time setting this problem up. I think I just need to use the
    d=rt but am not sure.

    Motorboat: current is 3mph. Going upstream takes 5 hours, going downstream takes 2.5 hours. What is the speed of the boat?
  • Feb 21, 2007, 09:44 AM
    Evil dead
    Speed = Distance / Time

    part a) If it takes 5 hours at a speed of 3mph. Then the distance traveled is 15 miles
    part b) If it takes 2.5 hours at a speed of 3mph. Then the distance traveled is 7.5 miles.

    That forms the foundation of your next calculation.

    What you are looking at is average speed over the distance traveled. The speed boat may be 3mph, but if it travels 5 hours then you are looking for an average speed, not the instantaneous speed which is where you went wrong.

    part a) Average speed = Instanteneous speed / Time taken
    Average speed = 0.6 mph

    part b) Average speed = Instanteneous speed / Time taken
    Average speed = 1.2 mph

    You can even error check by simply counting:

    part a) We need to make 5 hours at a speed of 0.6.
    At 0.6 PER hour (5hours x 0.6speed) we get 3.

    part b) We need to make 2.5 hours at a speed of 1.2
    At 1.2 PER hour (2.5hours x 1.2speed) we get 3.

    For any other calculations the distances we worked out should suffice ;-)
  • Feb 21, 2007, 09:46 AM
    Capuchin
    here you need to realise that the speed plus the current is twice as fast as thte speed minus the current

    I will use v for the speed and c for the current.

    so v + c = 2(v - c)

    v + c = 2v - 2c

    v = 3c = 9mph

    Hope you can understand this, please ask if you don't :)
  • Feb 21, 2007, 09:58 AM
    Evil dead
    Sorry my mistake...

    I don't understand what you did. Is that a standard rule? Or application of knowledge?
  • Feb 21, 2007, 10:00 AM
    Capuchin
    That was pretty much application of knowledge without using physics laws.

    I used the fact that the speed downstream is twice that upstream, formed equations for each direction in terms of boat speed and current, and then solved for boat speed.

    Because you aren't given a distance of the stream, I don't think you can use any other method.
  • Feb 21, 2007, 10:08 AM
    Evil dead
    Ahh I see.

    Time taken upstream is 5. Time taken downstream is 2.5

    So speed of river downstream is twice as much as upstream.

    Erm... what do you do next to be able to form the appropriate equation. So if I seem stupid, I'm trying my best these days to apply my knowledge..

    EDIT. Wouldn't the speed boat travel at the same speed as the river? Also, what happens to the 3mph speed of the current, where do you use it?
  • Feb 21, 2007, 10:16 AM
    Capuchin
    the boat is applying a constant force up and down the river

    the current is applying a force to the boat that is 3mph downstream

    when the boat is going upstream, it's total speed is v (it's speed) - c (the current)

    when the boat is going down stream, the total speed is v + c

    I realise that we should be working in forces and not speeds, but I don't think the OP is at that level of physics. So we are just dealing with speed. This is in the maths forum, so I suppose it's a maths question rather than using physical laws.

    now the boat upstream speed is half the speed down stream.

    upstream = 1/2downstream

    v-c = (v+c)/2

    then find v in terms of c

    put c in (3mph) at the end.
  • Dec 2, 2010, 03:31 AM
    tommmm
    Are you finished?
  • Dec 2, 2010, 10:34 AM
    Unknown008
    Quote:

    Originally Posted by tommmm View Post
    Are you finished?

    Yes, since 3 years ago.

  • All times are GMT -7. The time now is 06:29 AM.