Author Topic: Physics Help (Read 3159 times) Tweet Share

doboman · « **on:** December 25, 2007, 11:14:30 pm »

Hey guys, just wondering if anybody had jakaranda and know how to work out the first question in the first chapter.. im not sure how they got to that answer...any help will be nice

/0 · « **Reply #1 on:** December 26, 2007, 01:42:03 am »

Could you post the question?

Mao · « **Reply #2 on:** December 26, 2007, 10:06:32 am »

lol which year u in? that'll help!!
*searches for pdf but finds none*

Odette · « **Reply #3 on:** December 26, 2007, 10:37:20 am »

All I know is that he'll be in year 12 next year

hope that helps

doboman · « **Reply #4 on:** December 26, 2007, 11:45:32 am »

sorry about that, yes im in year 12..

Mao · « **Reply #5 on:** December 26, 2007, 12:53:18 pm »

no question??? plz post question rooboy!

doboman · « **Reply #6 on:** December 26, 2007, 05:45:51 pm »

Hey, sorry for the wait.
Q: Two physic students are trying to determine the instantanious speed of a bicycle 5.0m from the start of a 1000m sprint. They use a stopwatch to measure the time taken for the bicycle to cover the first 10m. if the acceleration is constant and the measured time is 4.0s what was the instantanious speen of the bicycle at the 5.om mark..

Ahmad · « **Reply #7 on:** December 26, 2007, 05:52:53 pm »

Edit: Wrong (misread, see posts below).

$x=5,\;u=0,\;t=4.0,\;v=?$

$x = \frac{u+v}{2} t \implies v = \frac{5}{2} \mbox{ms}^-1$

Mao · « **Reply #8 on:** December 26, 2007, 05:56:27 pm »

$ s=displacement=10m $

$ t=time=4s $

$ u=V_0=0 $

$ s=ut+\frac{1}{2}at^2 $
$ 10=\frac{1}{2}a \times 4^2 $
$ \frac{10}{8}=a $
$ a=1.25ms^{-2} $

at s=5:

$ v^2=u^2+2ad $

$ v^2=2 \times 1.25 \times 5 $

$ v=\sqrt{12.5} $

$ v \approx 3.5 ms^-1$

hope that helped

Collin Li · « **Reply #9 on:** December 26, 2007, 05:58:13 pm »

Ahmad is wrong.

$u=0\mbox{, }x=10\mbox{, }t=4\mbox{, }a=?$

$\mbox{Using }x = ut + \frac{1}{2}at^2$

$\Rightarrow 10 = 8a \Rightarrow a = \frac{5}{4} \mbox{ ms}^{-2}$

$\mbox{Now: }u = 0\mbox{, }a = \frac{5}{4}\mbox{, }x=5\mbox{, }v=?$

$\mbox{Using }v^2 = u^2 + 2ax$

$\Rightarrow v^2 = 2 \times \frac{5}{4} \times 5 = 12.5 \Rightarrow v = \sqrt{12.5} \approx 3.5\mbox{ ms}^{-1}$

Mao · « **Reply #10 on:** December 26, 2007, 06:05:50 pm »

Quote from: coblin on December 26, 2007, 05:58:13 pm

Ahmad is wrong.

ZOMG I R AT END OF ZE WORLD!!!!!!

Ahmad · « **Reply #11 on:** December 26, 2007, 06:08:20 pm »

Quote from: Obsolete Chaos on December 26, 2007, 06:05:50 pm

Quote from: coblin on December 26, 2007, 05:58:13 pm
Ahmad is wrong.
ZOMG I R AT END OF ZE WORLD!!!!!!

Not the first time, actually.

doboman · « **Reply #12 on:** December 26, 2007, 06:17:33 pm »

You guys rock man!! thanks so much

Eddict · « **Reply #13 on:** January 21, 2008, 08:35:57 pm »

I have a question, why did you substitute 5 into the second equation? Why not 10.. this is the problem I have with this..

midas_touch · « **Reply #14 on:** January 21, 2008, 08:49:41 pm »

Cos your finding the instantaneous speed when the bicycle has travelled 5m.

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