Way OT Speed conversion

[marc_scirocco at sympatico.ca (Mtl-Marc)]

> -----Message d'origine-----
> De : Erik Patterson [mailto:pats16v@columbus.rr.com] 
> Envoy? : 29 octobre 2006 20:19
> ? : Rocco List
> Objet : Way OT Speed conversion
> 
> Lurking mode off:
> 
> Hello all.  I have a question for the brains of the list.  
> How do I know the MPH of a given object knowing the weight, 
> amount of power, and time traveled a certain distance?  Any takers?
> 
> Six ounce object.
> 
> 8 Newtons of force
> 
> 90 feet traveled in 1.5 seconds
> 
> Yes this was a solid fuel rocket powered car.  It was 
> "modified" with the largest motors we could find on a sunday 
> in London Ohio.
> 
> Cheers,
> 
> Erik
>

What you are looking for is instantaneous speed which is the speed at a
particular moment in time.

We will assume that your rocket had a constant acceleration.

Predicting the position of an object undergoing constant acceleration
requires The formulae that we can derive from calculus concerning  constant
acceleration and motion.

"90 feet traveled in 1.5 seconds"

Metric conversion
90 feet = 27.432 meter
so,
27.432m in 1.5s

OK,
x = x0 + (v0 * t) + ((1/2) * a * (t*t))
where 
x = distance after object accelerates for a time t
x0 = initial distance
v0 = initial velocity
a = acceleration
t = time


27.432m = 0 + 0*0 + ((1/2) * a * (1.5s*1.5s))
27.432m = (1/2) * a * (1.5s*1.5s)
27.432m = (1/2) * a * 2.25s
a = 24.384m/sec*sec

To find the speed at 1.5s,
v = v0 + at
where
v = velocity (speed)
v0 = initial velocity
a = acceleration
t = time

v = 0 + (24.384m/sec*sec) * 1.5sec
v = 36.576m/s
36.576 meter/second = 131.67 kilometer/hour
or
36.576 meter/second = 81.81 mile/hour