Comments on: The Night Nothing Stood Still http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still Sassy O Level Maths Tuition, Questions & Tips from Singapore's Favourite Private Tutor Sun, 05 Feb 2012 17:08:39 +0000 hourly 1 By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24794 Miss Loi Thu, 30 Apr 2009 03:33:18 +0000 http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24794 <blockquote> As she turned to wave at the cheering crowd she'd totally forgotten about that 1st fallen tree at <var>s</var>=100 ... AND SHE SLAMMED INTO THE TREE 'GEORGE OF THE JUNGLE' STYLE! SOMEONE CALL THE AMBULANCE!!! </blockquote> *beeeeepoooooobeeeeepooooo*

As she turned to wave at the cheering crowd she'd totally forgotten about that 1st fallen tree at s=100 ... AND SHE SLAMMED INTO THE TREE 'GEORGE OF THE JUNGLE' STYLE!

SOMEONE CALL THE AMBULANCE!!!

*beeeeepoooooobeeeeepooooo*

]]> By: mathslover http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24761 mathslover Wed, 29 Apr 2009 12:33:13 +0000 http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24761 Time between Miss Loi passing point O and tree falling at point O = 10 - 9.87 = 0.13min = 7.8s. Speed of Miss Loi when she speed past point O = 763m/min = 45.78km/h. Can't imagine why Miss Loi's car would crawl under the tree when it looks to fall on her in 7.8 seconds' time. Anyway, now with a fallen tree at s=0 (i.e. point O) and another at s=100, with Miss Loi's car at 0<s<100, *surprise surprise* Miss Loi still chose to speed past s=100 at 810m/min!!! Unimaginable horrors... drama indeed. :) Time between Miss Loi passing point O and tree falling at point O = 10 - 9.87 = 0.13min = 7.8s.

Speed of Miss Loi when she speed past point O = 763m/min = 45.78km/h.

Can't imagine why Miss Loi's car would crawl under the tree when it looks to fall on her in 7.8 seconds' time.

Anyway, now with a fallen tree at s=0 (i.e. point O) and another at s=100, with Miss Loi's car at 0<s<100, *surprise surprise* Miss Loi still chose to speed past s=100 at 810m/min!!!

Unimaginable horrors... drama indeed. :)

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24750 Miss Loi Wed, 29 Apr 2009 06:44:18 +0000 http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24750 And now that the <a href="/questions/a-maths/the-night-nothing-stood-still#comment-24681" rel="nofollow">answers</a> to the question above are out, we have the following expressions for acceleration, velocity and displacement: <var>a</var> = 60<var>t</var> − 240 <var>v</var> = 30<var>t</var><sup>2</sup> − 240<var>t</var> + 210 <var>s</var> = 10<var>t</var><sup>3</sup> − 120<var>t</var><sup>2</sup> + 210<var>t</var> With this we can present a sketch of Miss Loi's <a rel="external nofollow" href="http://en.wikipedia.org/wiki/Initial_D" rel="nofollow">Initial D</a>-style route among the fallen trees ... complete with some F1-style commentary: <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2009/04/kinematics-motion-diagram.gif" alt="Kinematics Motion Diagram" /> <blockquote>And it begins with Miss Loi zooming past the starting line with a speed of 210 m/min ... Oh a fallen tree appears at 100m <em>after the start line</em>! ... She pulls her handbrake and turns the car around at <var>t</var> = 1 and passes the starting line again at 165m/min <em>in the opposite direction</em> ... Oh another fallen tree appears 980m <em>before the start line</em>! ... With a deft move, she spins the car around again at <var>t</var> = 1! ... and now she's reaching for the starting line again with the big tree that's gonna fall by <var>t</var> = 10! ... Can she make it?! Can she make it?!!! ... YES SHE PASSED THE STARTING LINE AT <var>t</var> = 9.87! Just listen to the cheers from the crowd! ... And she has zoomed away to 100m <em>after the start line</em> when the tree fell! Oh what drama!!!</blockquote> *pops champagne* <div class="attention"> While it is not required as part of your final working, it is always desirable to do a quick sketch of a motion diagram like the above to help you visualize the whole journey of the object in the question. This is especially useful when you're asked to calculate the total distance traveled (like part 3 of the question above) or average speed in the kinematics question </div> And now that the answers to the question above are out, we have the following expressions for acceleration, velocity and displacement:

a = 60t − 240
v = 30t2 − 240t + 210
s = 10t3 − 120t2 + 210t

With this we can present a sketch of Miss Loi's Initial D-style route among the fallen trees ... complete with some F1-style commentary:

Kinematics Motion Diagram

And it begins with Miss Loi zooming past the starting line with a speed of 210 m/min ... Oh a fallen tree appears at 100m after the start line! ... She pulls her handbrake and turns the car around at t = 1 and passes the starting line again at 165m/min in the opposite direction ... Oh another fallen tree appears 980m before the start line! ... With a deft move, she spins the car around again at t = 1! ... and now she's reaching for the starting line again with the big tree that's gonna fall by t = 10! ... Can she make it?! Can she make it?!!! ... YES SHE PASSED THE STARTING LINE AT t = 9.87! Just listen to the cheers from the crowd! ... And she has zoomed away to 100m after the start line when the tree fell! Oh what drama!!!

*pops champagne*

While it is not required as part of your final working, it is always desirable to do a quick sketch of a motion diagram like the above to help you visualize the whole journey of the object in the question. This is especially useful when you're asked to calculate the total distance traveled (like part 3 of the question above) or average speed in the kinematics question

]]> By: Li-sa http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24681 Li-sa Mon, 27 Apr 2009 10:20:32 +0000 http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24681 1. [pmath]v = int{}{}{a} dt = int{}{}{(60t - 240)} dt = 30t^2 - 240t + 210 [/pmath] ('.' when t = 0, v = 210 .'.constant of integration = 210) Speed at 2nd minute = [pmath]delim{|}{30(2)^2 - 240(2) + 210}{|} = 150[/pmath]m/min <div class="highlight"> <b>Miss Loi:</b> Yes the purpose of this first part is to let students be aware of the difference between <em>speed</em> (no minus sign possible) and <em>velocity</em> (minus sign possible). So performing the standard integration on the acceleration to get your velocity expression, we have: [pmath]v = int{}{}{a} dt = int{}{}{(60t - 240)} dt = 30t^2 - 240t + c[/pmath] And sub in <var>v</var>=210 when <var>t</var>=0 (when Miss Loi passed <var>O</var>) we get <var>c</var> = 210, so [pmath]v = 30t^2 - 240t + 210 [/pmath] --------- (1) Since the question asked for <em>speed</em>, and if the student doesn't have the habit of 'modding' || the expression what like <b>Li-sa</b> did, you'll get <var>v</var> = −150 m/min (after sub <var>t</var>=2) which is the <em>velocity</em>. This means that Miss Loi was actually travelling <em>backwards</em> at the second minute → so always remember to remove any − sign in your final answer when you're finding <em>speed</em> and <em>distance</em>! </div> 2. Turning back means that the speed becomes 0 m/min. .'. Solve [pmath]30t^2 - 240t + 210 = 0[/pmath] [pmath]30(t-1)(t-7) = 0[/pmath] t = 1 or 7 .'.Miss Loi has to turn back at t = 1 minute and t = 7 minutes. <div class="highlight"> <b>Miss Loi:</b> Those taking Physics would've had this ingrained in their minds - but it's vital to known that at each turning point, there will be this defining Zen-like 'Matrix' moment when everything stops, when there is <em>instantaneous rest</em> ... when <var>v</var> = 0. With this in mind, we can find the times at the turning points by equating the expression for <var>v</var> we've found in (1) to zero and solve for <var>t</var>, as <b>Li-sa</b> has done, and we find that Miss Loi had to turn back twice ('Matrix'-style) at the first and seventh minutes. </div> 3. [pmath]s = int{}{}{v} dt = int{}{}{(30t^2 - 240t + 210)} dt = 10t^3 - 120t^2 + 210t + 0[/pmath] ('.' at O, s = 0 .'. constant of integration = 0) When [pmath]t = 10, s = 10t^3 - 120t^2 + 210t = 0[/pmath] .'.The car will stop exactly where the tree fell. Miss Loi will not be able to pass it before it fell. <div class="highlight"> <b>Miss Loi:</b> Yes to know if Miss Loi managed to get pass the falling tree at <var>O</var> at <var>t</var> = 10, one way is to find out where she is at <var>t</var> = 10, and you'll also need to know what <em>direction</em> she's travelling from during this time. That's when a quick sketch like the one in the next comment below would be really helpful. Either way we'll need to obtain the expression for displacement <var>s</var> as shown by <b>Li-sa</b> above. Hence: [pmath]s = int{}{}{v} dt = int{}{}{(30t^2 - 240t + 210)} dt = 10t^3 - 120t^2 + 210t + c[/pmath] And sub in <var>s</var>=0 when <var>t</var>=0 (when Miss Loi passed <var>O</var>) we get <var>c</var> = 0, so [pmath]s = 10t^3 - 120t^2 + 210t[/pmath] --------- (2) So after substituting <var>t</var> = 10 into (2), we get <var>s</var>=100 ⇒ Miss Loi is 100m <em>after</em> point <var>O</var> (think there's a little careless mistake in Li-sa's working above). Also by substituting <var>t</var> = 10 into (1) Miss Loi's <em>velocity</em> is 810 m/min ⇒ she's traveling <em>forward</em> after <var>O</var> so she has definitely passed the falling tree by the tenth minute! *We can also equate (2) to zero and solve for <var>t</var>, we find that Miss Loi actually reached <var>O</var> earlier at <var>t</var> = 0, 2.13, 9.87. </div> Total distance she would have travelled in 10 minutes = [pmath]int{0}{10}{delim{|}{v}{|}}dt[/pmath] = [pmath]int{0}{10}{delim{|}{30t^2 - 240t + 210}{|}}dt[/pmath] = [pmath]int{0}{1}{(30t^2 - 240t + 210)}dt - int{1}{7}{(30t^2 - 240t + 210)}dt[/pmath] [pmath]+ int{7}{10}{(30t^2 - 240t + 210)}dt[/pmath] = 100 - (-1080) + 1080 = 2260 m <div class="highlight"> <b>Miss Loi:</b> It's important here for students to realize that merely substituting <var>t</var> = 10 into (2) will only give you the <em>displacement</em> from <var>O</var> at <var>t</var>=10. To get the total <em>distance</em>, you MUST add up the distances travelled at each parted of the journey segmented by the <em>turning points</em> obtained in Part 2 i.e. <var>t</var>=1 & 7. To do this, you can use <b>Li-sa</b>'s integration method above (note her +/&minus signs!) or note the displacements at each segment i.e. At <var>t</var> = 0, <var>s</var> = 0 At <var>t</var> = 1, <var>s</var> = 100 ⇒ total distance travelled from 0 to 1st min = 100 <var>t</var> = 7, <var>s</var> = −980 ⇒ total distance travelled from 1st to 7th min = 100+980 <var>t</var> = 10, <var>s</var> = 1000 ⇒ total distance travelled from 7th to 10th min = 980+100 ∴ Total distance travelled = 100+100+980+980+100 = 2260 m Again this will be much clearer if you've drawn a sketch of the journey (as shown below) ;) </div> Unless the tree falls the other way...:( 1. v = int{}{}{a} dt = int{}{}{(60t - 240)} dt = 30t^2 - 240t + 210
('.' when t = 0, v = 210 .'.constant of integration = 210)
Speed at 2nd minute = delim{|}{30(2)^2 - 240(2) + 210}{|} = 150m/min

Miss Loi: Yes the purpose of this first part is to let students be aware of the difference between speed (no minus sign possible) and velocity (minus sign possible).

So performing the standard integration on the acceleration to get your velocity expression, we have:

v = int{}{}{a} dt = int{}{}{(60t - 240)} dt = 30t^2 - 240t + c

And sub in v=210 when t=0 (when Miss Loi passed O) we get c = 210, so

v = 30t^2 - 240t + 210 --------- (1)

Since the question asked for speed, and if the student doesn't have the habit of 'modding' || the expression what like Li-sa did, you'll get v = −150 m/min (after sub t=2) which is the velocity. This means that Miss Loi was actually travelling backwards at the second minute

→ so always remember to remove any − sign in your final answer when you're finding speed and distance!

2. Turning back means that the speed becomes 0 m/min.
.'. Solve 30t^2 - 240t + 210 = 0
30(t-1)(t-7) = 0
t = 1 or 7
.'.Miss Loi has to turn back at t = 1 minute and t = 7 minutes.

Miss Loi: Those taking Physics would've had this ingrained in their minds - but it's vital to known that at each turning point, there will be this defining Zen-like 'Matrix' moment when everything stops, when there is instantaneous rest ... when v = 0.

With this in mind, we can find the times at the turning points by equating the expression for v we've found in (1) to zero and solve for t, as Li-sa has done, and we find that Miss Loi had to turn back twice ('Matrix'-style) at the first and seventh minutes.

3. s = int{}{}{v} dt = int{}{}{(30t^2 - 240t + 210)} dt = 10t^3 - 120t^2 + 210t + 0
('.' at O, s = 0 .'. constant of integration = 0)
When t = 10, s = 10t^3 - 120t^2 + 210t = 0
.'.The car will stop exactly where the tree fell. Miss Loi will not be able to pass it before it fell.

Miss Loi: Yes to know if Miss Loi managed to get pass the falling tree at O at t = 10, one way is to find out where she is at t = 10, and you'll also need to know what direction she's travelling from during this time. That's when a quick sketch like the one in the next comment below would be really helpful.

Either way we'll need to obtain the expression for displacement s as shown by Li-sa above. Hence:

s = int{}{}{v} dt = int{}{}{(30t^2 - 240t + 210)} dt = 10t^3 - 120t^2 + 210t + c

And sub in s=0 when t=0 (when Miss Loi passed O) we get c = 0, so

s = 10t^3 - 120t^2 + 210t --------- (2)

So after substituting t = 10 into (2), we get s=100 ⇒ Miss Loi is 100m after point O (think there's a little careless mistake in Li-sa's working above). Also by substituting t = 10 into (1) Miss Loi's velocity is 810 m/min ⇒ she's traveling forward after O so she has definitely passed the falling tree by the tenth minute!

*We can also equate (2) to zero and solve for t, we find that Miss Loi actually reached O earlier at t = 0, 2.13, 9.87.

Total distance she would have travelled in 10 minutes
= int{0}{10}{delim{|}{v}{|}}dt
= int{0}{10}{delim{|}{30t^2 - 240t + 210}{|}}dt
= int{0}{1}{(30t^2 - 240t + 210)}dt - int{1}{7}{(30t^2 - 240t + 210)}dt
+ int{7}{10}{(30t^2 - 240t + 210)}dt
= 100 - (-1080) + 1080
= 2260 m

Miss Loi: It's important here for students to realize that merely substituting t = 10 into (2) will only give you the displacement from O at t=10.

To get the total distance, you MUST add up the distances travelled at each parted of the journey segmented by the turning points obtained in Part 2 i.e. t=1 & 7.

To do this, you can use Li-sa's integration method above (note her +/&minus signs!) or note the displacements at each segment i.e.

At t = 0, s = 0

At t = 1, s = 100
⇒ total distance travelled from 0 to 1st min = 100

t = 7, s = −980
⇒ total distance travelled from 1st to 7th min = 100+980

t = 10, s = 1000
⇒ total distance travelled from 7th to 10th min = 980+100

∴ Total distance travelled = 100+100+980+980+100 = 2260 m

Again this will be much clearer if you've drawn a sketch of the journey (as shown below) ;)

Unless the tree falls the other way...:(

]]> By: rei99 http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24652 rei99 Sun, 26 Apr 2009 16:46:44 +0000 http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24652 ok..i give a better try next time..lol ok..i give a better try next time..lol

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24650 Miss Loi Sun, 26 Apr 2009 16:44:21 +0000 http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24650 <b>Krisandro:</b> Been thinking too much of botak <a href="http://www.krisandro.com/2009/04/22/ready-for-marriage/#comment-8855" rel="nofollow">babies with small eyes and fat noses</a> lately? <b>Rei:</b> 210 m/min = (210 × 60) m/hr = 12.6 km/hr ≠ 0.12km/hr Any slower and you'll have to come and help Miss Loi push the car! Krisandro: Been thinking too much of botak babies with small eyes and fat noses lately?

Rei: 210 m/min = (210 × 60) m/hr = 12.6 km/hr ≠ 0.12km/hr

Any slower and you'll have to come and help Miss Loi push the car!

]]> By: rei99 http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24648 rei99 Sun, 26 Apr 2009 16:14:45 +0000 http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24648 u speeding at 0.12km? (^_^) u speeding at 0.12km?

(^_^)

]]> By: krisandro http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24646 krisandro Sun, 26 Apr 2009 15:11:01 +0000 http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24646 To prevent hair from such incidents again. I suggest shaving it all off. "Diary of a BOTAK Private O Level Maths Tutor" Cool huh? To prevent hair from such incidents again. I suggest shaving it all off.

"Diary of a BOTAK Private O Level Maths Tutor"

Cool huh?

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24644 Miss Loi Sun, 26 Apr 2009 15:05:51 +0000 http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24644 Safety first! The roads were strewn with debris! Safety first! The roads were strewn with debris!

]]> By: FoxTwo http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24642 FoxTwo Sun, 26 Apr 2009 14:58:27 +0000 http://www.exampaper.com.sg/questions/a-maths/the-night-nothing-stood-still#comment-24642 Why are you driving so slow past the big tree anyway? 12km/h, people run also faster hehehehe! Why are you driving so slow past the big tree anyway? 12km/h, people run also faster hehehehe!

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