Avg Velocity vs. speed problem

Discussion in 'Computer Support' started by mart, Jan 28, 2010.

  1. mart

    mart Guest

    If one cycles 800M east for 4 mins, then stops for 4 mins, then cycles
    800M east for 2 mins, what is their avg velocity? Avg speed would be
    160M/Min (1600M/10mins) but since Velocity involves DIRECTION and
    speed would it be 1600M/6mins?
    mart, Jan 28, 2010
    #1
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  2. mart

    Whiskers Guest

    On 2010-01-28, mart <> wrote:
    > If one cycles 800M east for 4 mins, then stops for 4 mins, then cycles
    > 800M east for 2 mins, what is their avg velocity? Avg speed would be
    > 160M/Min (1600M/10mins) but since Velocity involves DIRECTION and
    > speed would it be 1600M/6mins?


    No.

    Average speed = distance divided by time

    Average velocity = displacement divided by time

    In your example, both are the same as direction doesn't change.

    But if that person had cycled 800 metres east in 4 minutes, stoppped for 2
    minutes, then cycled 800 metres /west/ in 4 minutes:

    Average speed = 1,600 metres divided by 10 minutes = 160 metres per
    minute.

    Average velocity = (800 metres east minus 800 metres west) divided by 10
    minutes = zero.

    <http://www.physics247.com/physics-homework-help/speed-velocity-acceleration.php>

    --
    -- ^^^^^^^^^^
    -- Whiskers
    -- ~~~~~~~~~~
    Whiskers, Jan 28, 2010
    #2
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  3. mart

    mart Guest

    On Thu, 28 Jan 2010 02:17:08 +0000, Whiskers wrote:

    > On 2010-01-28, mart <> wrote:
    >> If one cycles 800M east for 4 mins, then stops for 4 mins, then cycles
    >> 800M east for 2 mins, what is their avg velocity? Avg speed would be
    >> 160M/Min (1600M/10mins) but since Velocity involves DIRECTION and
    >> speed would it be 1600M/6mins?

    >
    > No.
    >
    > Average speed = distance divided by time
    >
    > Average velocity = displacement divided by time
    >
    > In your example, both are the same as direction doesn't change.
    >
    > But if that person had cycled 800 metres east in 4 minutes, stoppped for 2
    > minutes, then cycled 800 metres /west/ in 4 minutes:
    >
    > Average speed = 1,600 metres divided by 10 minutes = 160 metres per
    > minute.
    >
    > Average velocity = (800 metres east minus 800 metres west) divided by 10
    > minutes = zero.
    >
    > <http://www.physics247.com/physics-homework-help/speed-velocity-acceleration.php>


    Thanks, So both avg speed and velocity are the same in my example.
    mart, Jan 28, 2010
    #3
  4. mart

    Buffalo Guest

    mart wrote:
    > If one cycles 800M east for 4 mins, then stops for 4 mins, then cycles
    > 800M east for 2 mins, what is their avg velocity? Avg speed would be
    > 160M/Min (1600M/10mins) but since Velocity involves DIRECTION and
    > speed would it be 1600M/6mins?


    Very poorly worded. What is 800M for 4 mins?
    Do you mean the cyclist traveled 800meters in 4 minutes?
    Buffalo
    Buffalo, Jan 28, 2010
    #4
  5. mart

    mart Guest

    On Wed, 27 Jan 2010 21:33:30 -0700, Buffalo wrote:

    >
    >
    > mart wrote:
    >> If one cycles 800M east for 4 mins, then stops for 4 mins, then cycles
    >> 800M east for 2 mins, what is their avg velocity? Avg speed would be
    >> 160M/Min (1600M/10mins) but since Velocity involves DIRECTION and
    >> speed would it be 1600M/6mins?

    >
    > Very poorly worded. What is 800M for 4 mins?
    > Do you mean the cyclist traveled 800meters in 4 minutes?
    > Buffalo


    Yes, meters, should have used lowercase "m" and a space, however the
    distance unit is really irrelevant to the problem
    mart, Jan 28, 2010
    #5
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