Kinematics – This Is A-Maths Not Physics!

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Tuition given in the topic of A-Maths Tuition Questions from the desk of Miss Loi at 7:26 pm (Singapore time)

Updated on 14 March 2008

Admittedly, Miss Loi has been really busy this June holidays – from setting up The Temple to car accidents. So busy that she was unceremoniously deprived of her annual shopping pilgrimage during this period 🙁 .

Likewise, her army of devotees have also been unceremoniously starved of cerebral stimulation during this period. For this Miss Loi felt really bad, as she understands that no number of lame/bimbotic/scandalous posts on her personal life can ever substitute the mighty high you get when you’re attempting one of Miss Loi’s maths questions!

So to kickstart a new round of Maths problems, here’s one on kinematics that will help those of you who have just started to get acquainted with the topic these few weeks:

A particle moves in a straight line so that, at time t seconds after leaving a fixed point O, it’s velocity v m/s is given by

v = 6 – 5e^-(t/3)

Find the acceleration of the particle when t = 3.
Sketch the velocity-time graph for 0 ≤ t ≤ 4.
Find the displacement of the particle from O when t = 3.
Show that the particle will not pass through O again.

Detailed scientific analysis (by Miss Loi) has revealed a trend of such questions recurring in recent years i.e. it’s not just about kinematics anymore – differentiation, integration and curve sketching are usually thrown in for good measure.

CAUTION: If you’re a budding Einstein taking physics at the same time, there will be a dangerous tendancy for you to attempt these kind of questions using whatever super duper rocket science out-of-syllabus motion equations you’ve learnt in your physics class. Sadly, you WILL get confused – this is A-Maths NOT Physics! Don’t get embroiled needlessly in the eternal war between physics and pure mathematics – read 100 times the bold words in the last para and you’ll understand why! Think about how those who don’t take physics are going to answer the question.

That’s all Miss Loi has to say for now. Anyway, school’s in, fun’s out! Oh, get high now!

P.S. For some reason, Miss Loi finds the last part of the question a little sad … like this particle and O were never meant to be together *sob* (better cut down on those serial dramas!)

頑張って!!!

Miss Loi

kinematics

Revision Exercise

To show that you have understood what Miss Loi just taught you from the centre, you must:

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6 Comments

winston commented in tuition class
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Ok, anw have u tried the qns? 🙂
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Reply winston
Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class
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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.
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Li-sa commented in tuition class
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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. - 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 😉
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 = e^t curve 😉
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 .
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
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Reply Li-sa
Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class
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Li-sa, finally marked your workings. 你好勁喔!!!
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Reply Miss Loi
Li-sa commented in tuition class
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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
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Reply Li-sa
Li-sa commented in tuition class
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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)。
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Reply Li-sa