Results 1 to 3 of 3

Math Help - Motorboat Speed

  1. #1
    Banned
    Joined
    Jan 2007
    Posts
    315

    Motorboat Speed

    A motorboat heads upstream on a river that has a current of 3 mph. The trip upstream takes 5 hours, while the return trip takes 2.5 hours. What is the speed of the motorboat?

    Assume that the motorboat keeps a constant speed relative to the water.

    NOTE: What is the meaning of the assumption made above?

    I've seen upstream and downstream questions before. However, I always get confused when it comes to setting up the right table or chart that leads to the correct equation that leads to the answer.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,823
    Thanks
    317
    Awards
    1
    Quote Originally Posted by symmetry View Post
    A motorboat heads upstream on a river that has a current of 3 mph. The trip upstream takes 5 hours, while the return trip takes 2.5 hours. What is the speed of the motorboat?

    Assume that the motorboat keeps a constant speed relative to the water.

    NOTE: What is the meaning of the assumption made above?

    I've seen upstream and downstream questions before. However, I always get confused when it comes to setting up the right table or chart that leads to the correct equation that leads to the answer.
    The direction of flow of the water is opposite that of the motorboat for the first leg of the trip, so according to an observer on land the boat will move with a constant speed of v - 3 mph. Thus it takes 5 hours to cover an unknown distance of x miles, and follows the relationship:
    v - 3 = \frac{x}{5}

    On the return trip the boat covers the same distance x in a time of 2.5 hours, now moving with a speed of v + 3 according to a landlubber.
    v + 3 = \frac{x}{2.5} = \frac{x}{\frac{5}{2}} =  \frac{2x}{5} = 2 \cdot \frac{x}{5}

    (I've fiddled with the last equation to make it slightly more useful.)

    Now, solve the top equation for \frac{x}{5} (rather than for just x, simply for convenience. If you are unsure about this step, solve it for x.) Since this is already done, this is easy!
    \frac{x}{5} = v - 3

    Now insert this into the second equation:
    v + 3 = 2 \cdot (v - 3)

    v + 3 = 2v - 6

    -v = -9

    So v = 9 mph.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Jan 2007
    Posts
    315

    ok

    I never quiet solved for (x/5) but rather for x, y or z.

    However, your steps and explanation is very clear and simplistic.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Motorboat and Current
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 11th 2009, 04:54 AM
  2. final speed from escape speed
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: November 21st 2009, 01:23 PM
  3. Motorboat
    Posted in the Algebra Forum
    Replies: 4
    Last Post: December 7th 2008, 09:02 AM
  4. angular speed, tangetial speed.
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: November 24th 2007, 10:31 AM
  5. Converting horizontal speed to vertical speed
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: July 27th 2006, 06:23 AM

Search Tags


/mathhelpforum @mathhelpforum