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#### sxcalexc

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##### Motion Question
« on: January 28, 2008, 10:23:40 pm »
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Hey guys after looking at this question my first thoughts were to find the velocity during the 10th second where t=1 but I then realised that it's still accelerating during the second so you can't do v = d/t. I then though about simultaneous equations of motion but I don't think there are enough values. Care to enlighten me?

#### Collin Li

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##### Re: Motion Question
« Reply #1 on: January 28, 2008, 10:33:45 pm »
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It's accelerating uniformly, so you use the constant acceleration equations of motion. You know that it accelerates from rest ($u=0$) and that in the 10th second of motion, it moves 28.5 m, this means that:

$x_{10} - x_9 = 28.5$, where $x_t$ is the displacement from the starting point at any time t.

We can find $x_t$, with this formula:

$x = x_t = ut + \frac{1}{2}at^2$, but since $u=0$:

$\implies x_t = \frac{1}{2}at^2$

Therefore:

$x_{10} = 50a$ and $x_9 = 40.5a$

Using the above equation: $x_{10} - x_9 = 28.5$, therefore:

$\implies 50a - 40.5a = 28.5 \implies 9.5a = 28.5 \implies a = 3\mbox{ ms}^{-2}$
« Last Edit: January 28, 2008, 10:42:58 pm by coblin »

#### sxcalexc

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##### Re: Motion Question
« Reply #2 on: January 28, 2008, 10:41:32 pm »
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Cheers collin!

Ps. Love the Yoshi
« Last Edit: February 03, 2008, 06:30:35 pm by sxcalexc »

#### iamdan08

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##### Re: Motion Question
« Reply #3 on: March 12, 2008, 02:44:48 pm »
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Rather than start a new thread I'll post my motion question here:

A car travelling with a constant speed of $80km h^-^1$ passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating to $80km h^-^1$ in 10.0s and reaching a constant speed of $100km h^-^1$ after a further 5.0s. At what time will the policman catch up with the car?

Thanks!
2007-08 VCE - Accounting, Texts & Traditions, Methods, Chem, Physics, Lit

2011 Bachelor of Biomedicine (Completed) @ The University of Melbourne
2012 Doctor of Medicine (Second Year) @ The University of Melbourne

#### ed_saifa

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##### Re: Motion Question
« Reply #4 on: March 12, 2008, 03:01:44 pm »
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40 seconds?
« Last Edit: March 12, 2008, 03:03:30 pm by ed_saifa »
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#### iamdan08

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##### Re: Motion Question
« Reply #5 on: March 12, 2008, 03:17:51 pm »
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The back of the book says 32.5s.
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#### ed_saifa

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##### Re: Motion Question
« Reply #6 on: March 12, 2008, 03:27:30 pm »
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let me check my working.
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"It's not a community effort"
"It's not allowed. Only death is a valid excuse"
"Probably for the first time time this year I was totally flabbergasted by some of the 'absolute junk' I had to correct .... I was going to use 'crap' but that was too kind a word"
"How can you expect to do well when
-you draw a lemon as having two half-cells connected with a salt bridge
-your lemons come with Cu2+ ions built in" - Dwyer
"Why'd you score so bad?!" - Zotos

#### ed_saifa

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##### Re: Motion Question
« Reply #7 on: March 12, 2008, 03:30:09 pm »
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Did you try a velocity time graph?
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"It's not a community effort"
"It's not allowed. Only death is a valid excuse"
"Probably for the first time time this year I was totally flabbergasted by some of the 'absolute junk' I had to correct .... I was going to use 'crap' but that was too kind a word"
"How can you expect to do well when
-you draw a lemon as having two half-cells connected with a salt bridge
-your lemons come with Cu2+ ions built in" - Dwyer
"Why'd you score so bad?!" - Zotos

#### iamdan08

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##### Re: Motion Question
« Reply #8 on: March 12, 2008, 03:41:46 pm »
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Yeah i tried to sketch one but i wasn't sure what i was supposed to do from there, or whether you are able to use the equations of motion without drawing a graph.
2007-08 VCE - Accounting, Texts & Traditions, Methods, Chem, Physics, Lit

2011 Bachelor of Biomedicine (Completed) @ The University of Melbourne
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#### ed_saifa

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##### Re: Motion Question
« Reply #9 on: March 12, 2008, 03:47:48 pm »
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maybe thats why i keep getting the wrong answer. I kept drawing it on a velocity time graph.
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"It's not a community effort"
"It's not allowed. Only death is a valid excuse"
"Probably for the first time time this year I was totally flabbergasted by some of the 'absolute junk' I had to correct .... I was going to use 'crap' but that was too kind a word"
"How can you expect to do well when
-you draw a lemon as having two half-cells connected with a salt bridge
-your lemons come with Cu2+ ions built in" - Dwyer
"Why'd you score so bad?!" - Zotos

#### Collin Li

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##### Re: Motion Question
« Reply #10 on: March 12, 2008, 06:16:37 pm »
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Drawing the velocity-time graph should reveal:

1) The first 10 seconds shows the car overtaking the motorcycle. Assuming a constant acceleration, the distance gained (area between two curves) is:

$\left(80\mbox{ kmh}^{-1} \times \frac{1\mbox{ ms}^{-1}}{3.6\mbox{ kmh}^{-1}}\right) \times \frac{10}{2} = 111.1\mbox{ m}$

(I have converted the speed into SI units in this single step as well)

2) The next 5 seconds (t=10 to t=15) shows the motorcycle catching up to the car in distance (it's speed is now faster than the car). Assuming constant acceleration again, the distance caught up (area between two curves) is:

$\left(\left(100-80\mbox{ kmh}^{-1}\right) \times \frac{1\mbox{ ms}^{-1}}{3.6\mbox{ kmh}^{-1}}\right) \times \frac{5}{2} = 13.9\mbox{ m}$

So now, the car is only ahead of the motorcycle by 97.2 metres, and the car is catching up at a rate of $20\mbox{ kmh}^{-1} = 5.56\mbox{ ms}^{-1}$.

This means that it will take another $\frac{97.2\mbox{ m}}{5.56\mbox{ ms}^{-1}} = 17.5\mbox{ s}$ for the motorcycle to catch up.

This means the total time taken (including the initial 15 seconds we did manually) is $17.5 + 15 = 32.5\mbox{ s}$.
« Last Edit: March 12, 2008, 06:19:15 pm by coblin »

#### iamdan08

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##### Re: Motion Question
« Reply #11 on: March 12, 2008, 07:37:36 pm »
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Awsome! Now I understand, thanks heaps!
2007-08 VCE - Accounting, Texts & Traditions, Methods, Chem, Physics, Lit

2011 Bachelor of Biomedicine (Completed) @ The University of Melbourne
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#### Mao

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##### Re: Motion Question
« Reply #12 on: March 12, 2008, 07:57:49 pm »
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A more algebraic method:

speed of the car: $80kmh^{-1}=\frac{80}{3.6}ms^{-1}$

Distance by the car (in terms of t): $\frac{80}{3.6}\cdot t$

Acceleration of the police: $\frac{80}{3.6} \cdot \frac{1}{10s} = \frac{8}{3.6}ms^{-2}$

Time it took for the police to get to 100km/h: $\frac{100}{3.6} \div \frac{8}{3.6}=12.5s$

Distance by the police (in terms of t, partially using x=ut+0.5at2): $\frac{1}{2} \cdot \frac{8}{3.6} \cdot 12.5^2+\frac{100}{3.6}\cdot(t-12.5)$

equating these two functions in terms of t will arrive at the point when the distance covered are the same, i.e. the police has caught up with the motorcycle.
« Last Edit: March 12, 2008, 08:11:12 pm by Mao »
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