[AccessD] OT Friday - Calculus Problem

Thanks Doug.  I'll forward to him at school, maybe he'll get it off his
phone.  

Rocky


-----Original Message-----
From: accessd-bounces at databaseadvisors.com
[mailto:accessd-bounces at databaseadvisors.com] On Behalf Of Doug Steele
Sent: Friday, September 30, 2011 9:29 AM
To: Access Developers discussion and problem solving
Subject: Re: [AccessD] OT Friday - Calculus Problem

I can do it but I can't explain it well :)

if the position function is s(t) = -4.9t**2 + 200

then the function for the change of position in time (the velocity) is the
derivative of the first function (notice the apostrophe after the s).

s'(t) = -4.9 * 2 * t

solving for t= 4 gives you -39.2

Funny what you remember from high school!  Yes, I took calculus in high
school.  Newtonian calculus, at that, so I found university calculus a total
mind fzck.

Doug

On Fri, Sep 30, 2011 at 6:25 AM, Rocky Smolin <rockysmolin at bchacc.com>wrote:

> Dear Lists:
>
> My son has a calc test today and has a problem he doesn't understand.  
> I can't help him with it.  Anyone remember how to do this?
>
> Position function is s(t) = -4.9t**2 +200
>
> which gives the height in meters of an object that is falling from a 
> height of 200 meters. The velocity at time t = as seconds is given by:
>
> lim(t-->a) = ((s(a)-s(t)) / (a-t)
>
> FInd the velocity of the object when t=4. The answer in -39.2 m/sec.  
> How is that derived?
>
> MTIA
>
> Rocky
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