Comments on: Kinematics – This Is A-Maths Not Physics! http://www.exampaper.com.sg/questions/a-maths/kinematics-this-is-a-maths-not-physics Sassy O Level Maths Tuition, Questions & Tips from Singapore's Favourite Private Tutor Sun, 05 Feb 2012 17:08:39 +0000 hourly 1 By: Li-sa http://www.exampaper.com.sg/questions/a-maths/kinematics-this-is-a-maths-not-physics#comment-11089 Li-sa Sun, 27 Jul 2008 13:42:28 +0000 http://www.exampaper.com.sg/questions/a-maths/kinematics-this-is-a-maths-not-physics#comment-11089 我的同学都是「和鄰班斗」，我却想：不如跟整个世界斗吧。（不是因为香港的考试制度差，香港的考试制度一点也不差，是同学的要求低。） p.s.我是香港学生，来年考香港的会考(Hong Kong Certificate of Education Examination, 简称HK<b>CE</b>E)。 我的同学都是「和鄰班斗」，我却想：不如跟整个世界斗吧。（不是因为香港的考试制度差，香港的考试制度一点也不差，是同学的要求低。）

p.s.我是香港学生，来年考香港的会考(Hong Kong Certificate of Education Examination, 简称HKCEE)。

]]> By: Li-sa http://www.exampaper.com.sg/questions/a-maths/kinematics-this-is-a-maths-not-physics#comment-10472 Li-sa Wed, 09 Jul 2008 18:55:10 +0000 http://www.exampaper.com.sg/questions/a-maths/kinematics-this-is-a-maths-not-physics#comment-10472 不敢当不敢当。还是多学习的功夫。其实香港的中学会考（与你们的O-Levels差不多）绝对不会考这类的东西，以上的是香港高考的「數學及統計學」课程。 想来，我这个网上的香港交流生，才发现香港的学生是非常的幸运的，因为我们学的东西其实比别的国家容易。我之前踏进了神州大地，那里的学生学的也比香港的难。不但使课程，试想一想，香港、新加坡，都是弹丸之地，学生们的竞争对手也不多；但中国内地的学生，搏斗的是整个大国家的学生。 唉，我的同学何时才会明白他们「生在福中不知福」呢？ Li-sa, HK 不敢当不敢当。还是多学习的功夫。其实香港的中学会考（与你们的O-Levels差不多）绝对不会考这类的东西，以上的是香港高考的「數學及統計學」课程。

想来，我这个网上的香港交流生，才发现香港的学生是非常的幸运的，因为我们学的东西其实比别的国家容易。我之前踏进了神州大地，那里的学生学的也比香港的难。不但使课程，试想一想，香港、新加坡，都是弹丸之地，学生们的竞争对手也不多；但中国内地的学生，搏斗的是整个大国家的学生。

唉，我的同学何时才会明白他们「生在福中不知福」呢？

Li-sa, HK

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/kinematics-this-is-a-maths-not-physics#comment-10467 Miss Loi Wed, 09 Jul 2008 17:00:50 +0000 http://www.exampaper.com.sg/questions/a-maths/kinematics-this-is-a-maths-not-physics#comment-10467 <b>Li-sa</b>, finally marked your workings. 你好勁喔!!! Li-sa, finally marked your workings. 你好勁喔!!!

]]> By: Li-sa http://www.exampaper.com.sg/questions/a-maths/kinematics-this-is-a-maths-not-physics#comment-10314 Li-sa Thu, 03 Jul 2008 10:49:10 +0000 http://www.exampaper.com.sg/questions/a-maths/kinematics-this-is-a-maths-not-physics#comment-10314 1. Acceleration[pmath] {=} {dv}/{dt} = d/{dt} (6 - 5e^{-{t/3}})= 5/3 e^{-{t/3}}[/pmath] When t = 3, acceleration[pmath]{=} {dv}/{dt} = 5/3 e^{-{3/3}}[/pmath] = 0.61313 m/s² (cor. to 5 d.p.) <div class="highlight"> Acceleration is defined as the <strong>rate of change of velocity with time</strong> i.e. <m>{dv}/{dt}</m> - modified your workings (i.e. it's d<strong>t</strong> NOT dx) above for you ;) ) </div> 2. For 0 ≤ t ≤ 4, the y-intercept is 1 and there is no x-intercept. The curve is exponential but the slope decreases gradually. When t = 4, v = 4.7 (cor. to nearest 0.1). <div class="highlight"> A nice little diagram to illustrate what you've described ;) <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2008/07/kinematics-graph-answer.gif" alt="Answer Diagram" /> Since we're only interested in 0 ≤ <var>t</var> ≤ 4, we first sub in the values of <var>t</var> = 0, 4 into the equation to find the values of <var>v</var> at the extreme ends of the curve (i.e. <var>v</var> = 1 at <var>t</var> = 0 and <var>v</var> = 4.7 at <var>t</var> = 4. It's also worth noting that as <var>t</var> approaches infinity, <var>v</var> = 6. Lastly as to why is curve is shaped as such, the following diagram should show you how it's derived from the basic <var>v</var> = e<sup><var>t</var></sup> curve ;) <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2008/07/kinematics-graph-steps.gif" alt="Answer Diagram Steps" /> </div> 3.[pmath] s = int{0}{3}{(6 - 5e^{-{t/3}})}~dt = delim{[}{6t+15e^{-t/3}}{]} ^3 _0[/pmath] [pmath]{=} (6(3) + 15e^-1) - (6(0) + 15e^0)[/pmath] [pmath]{=} 18+15e^-1 - 15[/pmath] [pmath]{=} 3 + 15e^-1[/pmath] = 8.51819 m <div class="highlight"> Yup, the displacement <m>s = int {}{} v dt</m>. And do note that we will arrive at the same answer either via evaluating the <em>definite integral</em> as shown above i.e. with limits at <var>t</var> = 0, 3; or integrating the usual way taking into account the constant <var>c</var> i.e. <m>s = int {}{} {6 - 5e^{-t/3}} dt</m> <m>{} = 6t - {5e^{-t/3}}/{-1/3} + c</m> <m>{} = 6t + 15e^{-t/3} + c</m> When <var>t</var> = 0, particle is at O so <var>s</var> = 0, ⇒ 0 = 0 + 15e<sup>0</sup> + <var>c</var> ⇒ <var>c</var> = -15 ⇒ <m>s = 6t+15e^{-t/3}-15</m> So when <var>t</var> = 3, displacement from O <var>s</var> = 6(3)+15e<sup>-1</sup>-15 = 3+15e<sup>-1</sup> (<strong>SAME ANSWER AS ABOVE!</strong>) </div> 4.[pmath]lim{x right infty}{v} = lim{t right infty}{(6 - 5e^{-t/3})} = lim{t right infty}{(6 - 5/e^{t/3})} [/pmath] = 6 - 0 = 6 > 0 Since the particle moves in a straight line and the velocity does not approach 0 as x tends to infinity, the particle will not pass through O again. 若有纰漏，不吝赐教！ <div class="highlight"> Yes, in order for the particle to return to O, there must be a <em>turning point</em>. And if there's a turning point, the velocity <strong><var>v</var> is zero</strong> when it occurs. So as you've rightly pointed out, <var>v</var> will never, ever be zero at any point in time when <var>t</var> > 0 ⇒ no turning point ⇒ no turning back ⇒ never return to O! LOL </div> 1. Acceleration{=} {dv}/{dt} = d/{dt} (6 - 5e^{-{t/3}})= 5/3 e^{-{t/3}}
When t = 3, acceleration{=} {dv}/{dt} = 5/3 e^{-{3/3}} = 0.61313 m/s² (cor. to 5 d.p.)

Acceleration is defined as the rate of change of velocity with time i.e. {dv}/{dt} - modified your workings (i.e. it's dt NOT dx) above for you ;) )

2. For 0 ≤ t ≤ 4, the y-intercept is 1 and there is no x-intercept. The curve is exponential but the slope decreases gradually. When t = 4, v = 4.7 (cor. to nearest 0.1).

A nice little diagram to illustrate what you've described ;)

Answer Diagram

Since we're only interested in 0 ≤ t ≤ 4, we first sub in the values of t = 0, 4 into the equation to find the values of v at the extreme ends of the curve (i.e. v = 1 at t = 0 and v = 4.7 at t = 4.

It's also worth noting that as t approaches infinity, v = 6.

Lastly as to why is curve is shaped as such, the following diagram should show you how it's derived from the basic v = et curve ;)

Answer Diagram Steps

3.s = int{0}{3}{(6 - 5e^{-{t/3}})}~dt = delim{[}{6t+15e^{-t/3}}{]} ^3 _0
{=} (6(3) + 15e^-1) - (6(0) + 15e^0)
{=} 18+15e^-1 - 15
{=} 3 + 15e^-1 = 8.51819 m

Yup, the displacement s = int {}{} v dt.

And do note that we will arrive at the same answer either via evaluating the definite integral as shown above i.e. with limits at t = 0, 3; or integrating the usual way taking into account the constant c i.e.

s = int {}{} {6 - 5e^{-t/3}} dt
{} = 6t - {5e^{-t/3}}/{-1/3} + c
{} = 6t + 15e^{-t/3} + c

When t = 0, particle is at O so s = 0,
⇒ 0 = 0 + 15e0 + c
⇒ c = -15
⇒ s = 6t+15e^{-t/3}-15

So when t = 3, displacement from O

s = 6(3)+15e-1-15 = 3+15e-1 (SAME ANSWER AS ABOVE!)

4.lim{x right infty}{v} = lim{t right infty}{(6 - 5e^{-t/3})} = lim{t right infty}{(6 - 5/e^{t/3})}
= 6 - 0 = 6 > 0
Since the particle moves in a straight line and the velocity does not approach 0 as x tends to infinity, the particle will not pass through O again.
若有纰漏，不吝赐教！

Yes, in order for the particle to return to O, there must be a turning point. And if there's a turning point, the velocity v is zero when it occurs.

So as you've rightly pointed out, v will never, ever be zero at any point in time when t > 0
⇒ no turning point ⇒ no turning back ⇒ never return to O! LOL

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/kinematics-this-is-a-maths-not-physics#comment-427 Miss Loi Mon, 09 Jul 2007 16:42:52 +0000 http://www.exampaper.com.sg/questions/a-maths/kinematics-this-is-a-maths-not-physics#comment-427 Miss Loi just went in and saw many friends already solving your qns, so she decided not to spoil your party :) But you got Miss Loi's homepage address wrong! Tsk tsk. Miss Loi just went in and saw many friends already solving your qns, so she decided not to spoil your party :)

But you got Miss Loi's homepage address wrong! Tsk tsk.

]]> By: winston http://www.exampaper.com.sg/questions/a-maths/kinematics-this-is-a-maths-not-physics#comment-422 winston Sun, 08 Jul 2007 16:22:58 +0000 http://www.exampaper.com.sg/questions/a-maths/kinematics-this-is-a-maths-not-physics#comment-422 Ok, anw have u tried the qns? :) Ok, anw have u tried the qns? :)

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