No one knows how the legend of the Phantom Car of PIE began.

It is said that lousier drivers on the PIE every morning can sometimes hear the eerie squealing of tyres and the roar of an engine, accompanied by thumping techno music, that heralded the momentary appearance of a mysterious black car in their mirrors.

Eye witness accounts tell of a sexy slim lady with pale flawless complexion behind its wheel, with nary a spill from the paper cup of kopi peng perched on the cup-holder as she overtook from the left in one mesmerizing move, casting a bone-chilling glance of irritation as she passed.

During these moments, sleeping students being ferried to school in SUVs have been known to suddenly sit up and recite aloud trigonometry formulae, and taxi drivers have been reported to suddenly stop scolding the government and instead expound the applications of integration to their (disinterested) passengers.

All these, however, remain as an urban legend to a particular lorry driver cruising at a leisurely speed along the PIE’s right-most lane this morning, much to the annoyance of the long train of cars behind.

But it all changed when a vicious 100+ BPM techno track sounded from nowhere to drown out Teresa Teng‘s crooning voice in his stereo.

The needles in his instrument panel started to swing wildly, and all the denials in the world couldn’t prepare him from the blinding aura emanating from that innocuous-looking black car in his mirror …

Assuming the black car appeared right next to the lorry at t = 0, the diagram shows the speed-time graph of the lorry and the black car during a period of 60 seconds.

Find

the acceleration of the car in the first 10 seconds;

the distance travelled by the car during the 60 seconds;

the distance travelled by the lorry during the 60 seconds;

the time when the car overtakes the lorry (if it does), and the time when the lorry overtakes the car (if it does).

NOTE: When you see a speed-time graph in an E-Maths Kinematics question, it’s almost inevitable that you’ll have to recall that:

Distance covered = area under the graph

DON’T be confused with your average speed = total distance/total time formula you learnt in PSLE!

P.S. While many of Miss Loi’s Sec 3 students were smirking at her for insulting their intelligence when doing Parts 1-3, for some reason quite a number of them weren’t smirking anymore when they reached Part 4.

Thought was some urban legend haunted story, turned out to be a such an interesting math question. Lol! Some school or assesment book publisher should get Miss Loi to set questions. Nice one!

Pacey: Think Miss Loi's assessment books will be mighty thick if she gets to set the questions since they're so chiong hei lol

Cockcroach: Students need to be able to recite the formulae first BEFORE they can chant "WE WILL SCORE A1s!" i.e. have to learn to walk first before flying 😉

Krisandro: You don't like Techno? Minus 100 marks! F9 for you!

*Casts a bone-chilling glance at Soupie*

*Casts another bone-chilling glance at Huifen*

Clarion: Got. Depends on which school you're from 😉

Miss Loi, I have attempted parts 1 to 4 of your question. Part 1: Acceleration = {20-0}m/s/{10-0}s = 2 m/s^{2}

Yes in a linear speed-time graph, the acceleration (=rate of change of speed) over a period of time is the gradient of the graph within this period. But remember to include the unit in you answer (see above)!

P.S. It's also worth noting that when the gradient is negative (i.e. at the t=40-60s period in the graph), the car is decelerating i.e. slowing down.

Part 2: Distance travelled by the car = 1/2*((60-0)+(40-10))*20 = 900 m

Yes as mentioned, the distance covered in a speed-time graph will be the area under the curve which is a trapezium for the range of 0-60s. So using the area of trapezium formula (i.e. 1/2 * sum of parallel sides * height) you'll get the total distance of 900m.

For those who've somehow forgotten the trapezium area formula, you can also add up the areas of the triangle (t=0-10s), rectangle (t=10-40s), and triangle (t=40-60s) and you should get the same answer.

Such a peanut question right?

Part 3: Distance travelled by the lorry = 15*60 = 900 m

Miss Loi will knock your head if you don't know how to find the area of the rectangle for t=0-60s! But beware of careless mistakes here as students can sometimes get double-vision and confused as to which graph are they dealing with now - we're dealing with the lorry not the car in this part!

This is also the time when Miss Loi's students start smirking. Tsk tsk.

For part 4, I plotted a graph on computer. I derived some equations that fit the given speed-time graph, then I integrated them. Hope it does not matter. The graph is titled "pix1" in my public photo album.

Many apologies for the crude "formulas" above. Hope they do not cause much readability problems. I finally derived the algebraic solution to part 4, as follows:

Let t s be the required time for the car to be at the same position as the lorry. .'.

.'. When the time is just over 20 s, the car overtakes the lorry. As both vehicles arrive at the same location when time = 60 s, the lorry does not actually overtake the car. Illustrated in the graph.

When I saw part 4 of this question, my instinct was not to use an equation, thus I plotted a graph. However the algebra did not seem anywhere obvious in the graph, so I resorted to using the equation. 无奈，但很有用。

Was about to say that graphical calculators are still banned in O-Level exams so graphing solutions are out - even though fggffggf's answers are correct.

In order to solve the smirk-erasing Part 4, one has to understand that:

At the time when the car overtakes the lorry (and vice versa), they are at the same position i.e. their distance traveled since t = 0 is THE SAME.

*Assuming that both car and lorry started at the same position when t = 0 (Miss Loi added this in the question but sometimes they don't specify - it's good to state this assumption in your working)

So to find the overtaking time, you'll need to express the distance travelled by the car and lorry in terms of t, equate them and solve for t.

A. Distance traveled by lorry in time t = 15t ----- (1)

B. You can express the distance traveled by car in time t straightaway using the trapezium area formula as shown by fggffggf above OR do it part by part just to check that you're on the right track:

As can be seen from the diagram:

When t = 10, distance covered by car < distance covered by lorry ⇒ car hasn't yet overtaken lorry.

When t = 40, distance covered by car > distance covered by lorry ⇒ car has overtaken lorry.

⇒ car overtook lorry somewhere 10 < t < 40

∴ distance expression for car = total distance traveled in 1st 10 sec + distance traveled for 0 < t < 40 ⇒ 100 + (t - 10)(20) ----- (2)

Equate (1) & (2) and solving you'll get t = 20s 😉

As to whether the lorry overtakes the car again, the answers in Parts 2 and 3 already show that both of them traveled the same distance in 60 sec, so the lorry did overtake the car again at t = 60s because the lorry is still continuing at a constant speed while the car has come to a stop (presumably by more traffic jam on her lane!)

Oh and thanks fggffggf for contributing your answers. Great example of multiple approaches to the same question here - though graphical calculators are still banned for the O-Levels!

Clarion: If you're a Physics student, as mentioned previously, please refrain from thinking about those super duper rocket science out-of-syllabus motion equations you've learnt in your physics class when tackling your Maths kinematics questions - you risk getting yourself mightily confused in the end!

It is good reminding students the ban of graphing calculators. (Actually I am not a Singaporean student so I did not know that, but it does not matter; many thanks to pointing that I missed the units in part 1.)

Omg Miss Loi is so pretty. Haha. I was browsing the papers when I noticed a sexy young math tutor on the front page, and I know it's you. Hahahaha. It's a pretty nice pretty photo. 🙂

* Please refer to the relevant Ten-Year Series for the questions of these suggested solutions.

About Miss Loi

Miss Loi is a full-time private tutor in Singapore specializing in O-Level Maths tuition. Her life’s calling is to eradicate the terrifying LMBFH Syndrome off the face of this planet. For over years she has been a savior to countless students … [read more]

Gratitudes

Frankly, I was amazed by your ability to make an F9 student like me, to become an A1 student within months … [read more]

## 16 Comments

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Thought was some urban legend haunted story, turned out to be a such an interesting math question. Lol! Some school or assesment book publisher should get Miss Loi to set questions. Nice one!

曜

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Aren't the students reciting, "WE WILL SCORE A1s..."?

曜

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You listen to techno?

*minus 20 points off the sexeh math teacher*

曜

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Hi sexy slim lady..haha..

曜

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omg.. i thought its a horror thingy.. seems to be quite interesting.. who knows... end up with maths.. **sien** my maths sux la weh..

曜

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sec 3 gt learn accel. time graph meh?

曜

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Pacey:Think Miss Loi's assessment books will be mighty thick if she gets to set the questions since they're sochiong heilolCockcroach:Students need to be able to recite the formulae firstBEFOREthey can chant "WE WILL SCORE A1s!" i.e. have to learn to walk first before flying 😉Krisandro:You don't like Techno? Minus 100 marks! F9 for you!*Casts a bone-chilling glance at

Soupie**Casts another bone-chilling glance at

Huifen*Clarion:Got. Depends on which school you're from 😉曜

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Miss Loi, I have attempted parts 1 to 4 of your question.

Part 1: Acceleration = {20-0}m/s/{10-0}s = 2

m/s^{2}Yes in a linear speed-time graph, the acceleration (=rate of change of speed) over a period of time is the

gradientof the graph within this period. But remember to include the unit in you answer (see above)!P.S. It's also worth noting that when the gradient is

negative(i.e. at thet=40-60s period in the graph), the car isdeceleratingi.e. slowing down.Part 2: Distance travelled by the car

= 1/2*((60-0)+(40-10))*20

= 900 m

Yes as mentioned, the distance covered in a speed-time graph will be the area under the curve which is a trapezium for the range of 0-60s. So using the area of trapezium formula (i.e. 1/2 * sum of parallel sides * height) you'll get the total distance of 900m.

For those who've somehow forgotten the trapezium area formula, you can also add up the areas of the triangle (

t=0-10s), rectangle (t=10-40s), and triangle (t=40-60s) and you should get the same answer.Such a peanut question right?

Part 3: Distance travelled by the lorry

= 15*60 = 900 m

Miss Loi will knock your head if you don't know how to find the area of the rectangle for

t=0-60s! But beware of careless mistakes here as students can sometimes get double-vision and confused as to which graph are they dealing with now - we're dealing with the lorry not the car in this part!This is also the time when Miss Loi's students start smirking. Tsk tsk.

For part 4, I plotted a graph on computer. I derived some equations that fit the given speed-time graph, then I integrated them. Hope it does not matter.

The graph is titled "pix1" in my public photo album.

Oops class starting ... will be right back 😛

曜

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Many apologies for the crude "formulas" above. Hope they do not cause much readability problems.

I finally derived the algebraic solution to part 4, as follows:

Let

ts be the required time for the car to be at the same position as the lorry..'.

.'. When the time is just over 20 s, the car overtakes the lorry.

As both vehicles arrive at the same location when time = 60 s, the lorry does not actually overtake the car. Illustrated in the graph.

When I saw part 4 of this question, my instinct was not to use an equation, thus I plotted a graph. However the algebra did not seem anywhere obvious in the graph, so I resorted to using the equation. 无奈，但很有用。

曜

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reminds me of physics...

曜

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Ok it has been a looong day ... 😀

Was about to say that graphical calculators are still banned in O-Level exams so graphing solutions are out - even though

fggffggf's answers are correct.In order to solve the smirk-erasing Part 4, one has to understand that:

At the time when the car overtakes the lorry (and vice versa), they are at the same position i.e. their distance traveled since

t= 0 is THE SAME.*Assuming that both car and lorry started at the same position when

t= 0 (Miss Loi added this in the question but sometimes they don't specify - it's good to state this assumption in your working)So to find the overtaking time, you'll need to express the distance travelled by the car and lorry in terms of

t, equate them and solve fort.A. Distance traveled by lorry in time

t= 15t----- (1)B. You can express the distance traveled by car in time

tstraightaway using the trapezium area formula as shown byfggffggfabove OR do it part by part just to check that you're on the right track:As can be seen from the diagram:

t= 10, distance covered by car < distance covered by lorry ⇒ car hasn't yet overtaken lorry.t= 40, distance covered by car > distance covered by lorry ⇒ car has overtaken lorry.t< 40∴ distance expression for car = total distance traveled in 1st 10 sec + distance traveled for 0 <

t< 40⇒ 100 + (

t- 10)(20) ----- (2)Equate (1) & (2) and solving you'll get

t= 20s 😉As to whether the lorry overtakes the car again, the answers in Parts 2 and 3 already show that both of them traveled the same distance in 60 sec, so the lorry did overtake the car again at

t= 60s because the lorry is still continuing at a constant speed while the car has come to a stop (presumably by more traffic jam on her lane!)曜

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Oh and thanks

fggffggffor contributing your answers. Great example of multiple approaches to the same question here - though graphical calculators are still banned for the O-Levels!Clarion:If you're a Physics student, as mentioned previously, please refrain from thinking about those super duper rocket science out-of-syllabus motion equations you've learnt in your physics class when tackling yourMathskinematics questions - you risk getting yourself mightily confused in the end!This is Maths NOT Physics!

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@Miss Loi: Do physic simpler leh... i see amath kinematics and its 3x as hard as physics XP

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It is good reminding students the ban of graphing calculators. (Actually I am not a Singaporean student so I did not know that, but it does not matter; many thanks to pointing that I missed the units in part 1.)

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Hi Miss Loi,

Congratulations for having been featured in The Straits Times today as being one of five private tutors on the front page....cool.

Ordinary Guy

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Omg Miss Loi is so pretty. Haha. I was browsing the papers when I noticed a sexy young math tutor on the front page, and I know it's you. Hahahaha. It's a pretty nice pretty photo. 🙂