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#### TrueTears

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« on: January 12, 2009, 11:55:27 pm »
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soz, I think i might be posting too much threads for just 1 question, so i'll make a separate thread XD (please excuse my noob questions haha). Here goes...

Two physics students are trying to determine the instantaneous speed of a bicycle 5m from the start of a 1000m sprint. They use a stop watch to measure the time taken for the bicycle to cover the first 10m. If the acceleration was constant and the measured time was 4s. What was the instantaneous speed of the bicycle at the 5 m mark?
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#### mullums1

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« Reply #1 on: January 13, 2009, 12:01:20 am »
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x= 10
t=4
u=0
a=?

Find "a" using x= u + .5 at^2

then after finding "a", find v using v= u + at
(sub in your calculated "a" value, t= 4 secs and U= 0)

#### TrueTears

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« Reply #2 on: January 13, 2009, 12:04:09 am »
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yeah thanks XD i just didnt know wat that 1000m sprint meant, why did they even put that info? lol
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#### mullums1

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« Reply #3 on: January 13, 2009, 12:05:42 am »
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yeah thanks XD i just didnt know wat that 1000m sprint meant, why did they even put that info? lol

Yeh, they will try to put you off with extra info. Just try to extract the relevation info and assosicate that with the right formula.

#### Flaming_Arrow

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« Reply #4 on: January 13, 2009, 12:07:31 am »
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$d = 10 t = 4 u=0$

$d = ut + \frac {at^2}{2}$

$10 = 8a$

$a = \frac {5}{4} ms^{-2}$

$u=0 a=\frac {5}{4} d = 5$

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

$v^2 = 2 * \frac {5}{4} * 10$

$v = 3.5 ms^{-1}$
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#### vce08

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« Reply #5 on: January 13, 2009, 12:07:49 am »
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hey hey
You are starting Physics (my favourite subject in the VCE)

#### Flaming_Arrow

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« Reply #6 on: January 13, 2009, 12:10:08 am »
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x= 10
t=4
u=0
a=?

Find "a" using x= u + .5 at^2

then after finding "a", find v using v= u + at
(sub in your calculated "a" value, t= 4 secs and U= 0)

after you find a you can't sub in t = 4 because that will mean he travelled 10m, we r trying to find speed when its 5m
2010: Commerce @ UoM

#### TrueTears

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« Reply #7 on: January 13, 2009, 12:12:00 am »
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hey hey
You are starting Physics (my favourite subject in the VCE)

haha thanks Table, gonna be fun XD
PhD @ MIT (Economics).

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#### TrueTears

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« Reply #8 on: January 13, 2009, 02:56:10 pm »
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An old light globe hangs by a wire from the roof of a train. What angle does it make with the vertical when the train is accelerating at $1.5ms^-2$
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#### Flaming_Arrow

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« Reply #9 on: January 13, 2009, 02:58:02 pm »
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hint: use $g = 10 ms^{-2}$
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#### vce08

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« Reply #10 on: January 13, 2009, 03:11:59 pm »
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Pythagoras i believe

#### danieltennis

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« Reply #11 on: January 13, 2009, 03:29:44 pm »
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Horizontal component $= 1.5 ms^{ - 2}$     Vertical Component = $10 ms^{ - 2}$

$tan\theta = 1.5/10$
$\theta = tan ^{ - 1}(1.5/10)$
$\theta = 8.5 ^\circ$
« Last Edit: January 13, 2009, 04:09:10 pm by daniel. »

#### /0

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« Reply #12 on: January 13, 2009, 04:02:42 pm »
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An old light globe hangs by a wire from the roof of a train. What angle does it make with the vertical when the train is accelerating at $1.5ms^-2$

Dealing with accelerating reference frames can be tricky. I found the most intuitive way to go is to let the tension in the rope be $T$, the mass of the globe be $m$, and let the angle between the rope and the vertical be $\theta$.

The force $mg$ acts downwards. The force $1.5m$ acts to the right. Therefore to balance those forces,

$T\sin\theta = 1.5m$

$T\cos\theta=mg$

$\therefore \tan\theta = \frac{1.5}{g}$, by division.

#### TrueTears

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« Reply #13 on: January 13, 2009, 07:00:57 pm »
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thanks

also:
1. A railway cart of 500kg filled with coal weighing 250kg moves at a speed of 2$ms^-1$. But suddenly the entirely load of coal falls out from the railway cart through a hole in the cart. What is the final velocity of the cart?

i did 2(250 + 500) = 500v + 250 v

v = 2ms^-1 . which is correct answer but how do u know that the coal and the cart travels at the same final speed?
« Last Edit: January 13, 2009, 07:36:21 pm by TrueTears »
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#### Flaming_Arrow

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$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$