James,

The general form is
d = v0 * t + 0.5 * a * t^2

You have added the problem of a maximum speed.  To overcome this
find
t1 = (vf - v0) / a
t2 = t - t1

if t1 is greater than t (you never hit maximum speed), use the
above formula.  Otherwise use
d = v0 * t1 + 0.5 * a * t1^2 + vf * t2

In your example,
t = 1 second
v0 = 2 inches / sec
vf = 20 inches / sec
a = 30 inches / sec^2 (right?)

t1 = (20 - 2) / 30 = 0.6 sec
t2 = 1 - 0.6 = 0.4 sec
d = 2 * 0.6 + 0.5 * 30 * 0.6^2 + 20 * 0.4 = 14.6 inches

definitions:
t = total time
t1 = time to hit final speed
t2 = time at final speed
d = distance
v0 = initial speed
vf = final speed
a = acceleration

Quoting James Williams <jlw.creditview@VERIZON.NET>:

> Hello,
>
> Does anyone know how to calculate distance moved give a starting
> speed,
> linear acceleration, time and ending speed?  I am wanting to figure out
> how
> are I have travel in distance during my acceleration up to full speed.
>
> I.E.  If I start motion at 2inches/sec and I have a acceleration of
> 30inches/sec and my top speed is 20inches/sec and my time is 1sec, how
> far
> have I actually moved?
>
>
> Regards,
>
> James
>

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