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Posted

If I know the distance an object travels from point A to point B, that it starts from a stopped position and the time that it takes to get from point A to point B, how can I calculate the top speed at point B?

Posted

You can calculate the average speed using

 

[math] speed = \frac{distance}{time \:\:taken}[/math].

 

Without knowing any more we can't calculate things like instantaneous speed or the maximum speed.

Posted (edited)

dude, s=at^2

 

or, rearranged a=s/t^2

 

where a = acceleration

t=time

s= displacement.

 

we do know the acceleration.

Edited by insane_alien
Posted
dude, s=at^2 or, rearranged a=s/t^2

 

Dude, you missed a factor of 1/2: [math]d=\frac 1 2 at^2[/math] (I prefer 'd' for displacement; using 's' is confusing as it is also used for speed.) Combining with [math]v=at[/math] and solving for the final velocity in terms of distance and time,

 

[math]v=at = 2\frac d t[/math]

Posted (edited)

Surely, with constant acceleration, the maximum velocity is twice the average velocity (assuming you start at zero velocity)? It's a straight line graph of velocity against time. Or am I missing something?

 

989 / 27.009 = 36.6174 ft/sec ave. velocity

So Max velocity = 36.6174 * 2 = 73.235 ft/sec.

 

I'll let you do the conversion to mph.

Edited by jedaisoul
Posted
DH just said that

Yes, I did not spot the significance of the last line of DH's post. But then again, I wonder whether the enquirer would have grasped the meaning of the formal proof given? My solution was less rigorous, but I hope more intelligible to a casual reader.

  • 3 weeks later...
Posted

x = xo + vot + at²/2

989 = 0 + 0.t + a(27,009)²/2

989 = a(27,009)²/2

989 . 2 = a(729,486081)

1978 = a (729,486081)

a = 1978/729,486081

a = 2,7114979319255852943409347929697

 

in feet/second

just convert :D

 

xD

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