Re: speed distance & time

["Dan" <fordghia@ozemail.com.au>]

Hubert Lam <hubert@imap4.com> wrote in message
95kvhscojq2gflt6gu8t7k51ck148iqbjb@4ax.com">news:95kvhscojq2gflt6gu8t7k51ck148iqbjb@4ax.com...

Hubert - It is about time we should discuss Physics as I'm trying to work
out the average constant speed (and time) as the train moves and stops from
Turella to Central via Sydenham and Turella to Central via Airport Link. But
one thing missing - I can't have access to the speedometer on the train.

>           i am just writing to ask about a mathematical formula.
>           i am trying to work out how fast an object travels and the other
part (i
>           think if its done backwards it tells you how long it took), but
im not to
>           sure how it goes i know that Speed Time and Distance is
involved.
>
> v = s/t
>
> where v is velocity (or speed if you don't care about direction)
> s is displacement (or distance if you don't care about direction)
> and t is time.
>
> from this you can derive the following:
>
> v = u + at
>
> where a is the acceleration of the object
> v is the final velocity
> and u is the initial velocity
>
> and hence you have
>
> v^2 = u^2 + 2as
>
> and s = ut + 0.5a(t^2)
>
> All you need to know is three of the variables and find the relevant
equation of motion and plug
> (substitute) the numbers into the correct places.
>
>
>