pworkmaster7336 Posted August 2, 2008 Share Posted August 2, 2008 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? Link to comment Share on other sites More sharing options...
ajb Posted August 2, 2008 Share Posted August 2, 2008 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. Link to comment Share on other sites More sharing options...
insane_alien Posted August 2, 2008 Share Posted August 2, 2008 you could assume a constant acceleration and work it out from there as it does say it started at 0 velocity. Link to comment Share on other sites More sharing options...
swansont Posted August 2, 2008 Share Posted August 2, 2008 If you do have constant acceleration, you would use v = at (since it started at rest) Link to comment Share on other sites More sharing options...
pworkmaster7336 Posted August 2, 2008 Author Share Posted August 2, 2008 Assuming the object has constant acceleration and travels 989 feet in 27.009 seconds, what is the top speed in mph? Link to comment Share on other sites More sharing options...
Fuzzwood Posted August 2, 2008 Share Posted August 2, 2008 Impossible to say without knowing the acceleration. Only thing you can calculate from this is Vavg Link to comment Share on other sites More sharing options...
insane_alien Posted August 2, 2008 Share Posted August 2, 2008 (edited) dude, s=at^2 or, rearranged a=s/t^2 where a = acceleration t=time s= displacement. we do know the acceleration. Edited August 2, 2008 by insane_alien Link to comment Share on other sites More sharing options...
D H Posted August 2, 2008 Share Posted August 2, 2008 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] Link to comment Share on other sites More sharing options...
insane_alien Posted August 2, 2008 Share Posted August 2, 2008 gah. i've been doing that a lot lately. think i'll go take a nap. Link to comment Share on other sites More sharing options...
jedaisoul Posted August 8, 2008 Share Posted August 8, 2008 (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 August 8, 2008 by jedaisoul Link to comment Share on other sites More sharing options...
insane_alien Posted August 8, 2008 Share Posted August 8, 2008 DH just said that Link to comment Share on other sites More sharing options...
jedaisoul Posted August 10, 2008 Share Posted August 10, 2008 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. Link to comment Share on other sites More sharing options...
Severo Posted August 30, 2008 Share Posted August 30, 2008 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 xD Link to comment Share on other sites More sharing options...
pworkmaster7336 Posted August 31, 2008 Author Share Posted August 31, 2008 :doh:Thanks for your help! I think I've got it now. Link to comment Share on other sites More sharing options...
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