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 mart 01-28-2010 12:23 AM

Avg Velocity vs. speed problem

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?

 Whiskers 01-28-2010 02:17 AM

Re: Avg Velocity vs. speed problem

On 2010-01-28, mart <mxxx@nowherexcv.net> 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
-- ~~~~~~~~~~

 mart 01-28-2010 04:29 AM

Re: Avg Velocity vs. speed problem

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

> On 2010-01-28, mart <mxxx@nowherexcv.net> 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.

 Buffalo 01-28-2010 04:33 AM

Re: Avg Velocity vs. speed problem

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

 mart 01-28-2010 05:35 AM

Re: Avg Velocity vs. speed problem

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

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