26[A Math CD]

1 © Penerbitan Pelangi Sdn. Bhd.

Paper 2

1. Gradient of the graph = rate of change of temperature

2. Average speed = gradient of the graph

= 20–––4

= 5 m s−1

3. Rate of change of speed = gradient of the graph

= 28–––8

= 3.5 m s−2

4. Rate of change of speed = gradient of the graph

= 12 – 6––––––0 – 4

=−1.5ms−2

5. Average speed = gradient of the graph

= (10−2)km–––––––––––30 minutes

= 8km––––––––0.5 hour

=16kmh−1

6. (a) Rateofchangeofspeedinthefirst5s

= 18−0––––––5−0

= 3.6 m s−2

(b) Rateofchangeofspeedinthelast4s

= 0 – 18––––––9 – 5

=−4.5ms−2

7. Distancetravelled=areaunderthegraph = 8 × 6 = 48 m

8. Distancetravelled=areaunderthegraph

= 1—2

× 8 × 7

= 28 m

9. Distancetravelled =areaunderthegraph

= 1—2

×(6+10)× 9

= 72 m

Paper 2

1. (a) Averagespeed

= 400.521

= 19.07 m s–1


= 0 – 27–––––––21 – 12

=−3ms−2

(c) 12

× (u+27)×5+(12–5)× 27

+ 12

×(21–12)× 27 = 400.5

2.5u + 67.5 + 189 + 121.5 = 400.5 2.5u = 400.5 – 121.5

−189−67.5

u = 22.5––––2.5

= 9

2. (a) Length of time that the particle moves withuniform speed

=17−5 = 12 seconds


= 30 – 017 – 32

= –2 m s−2

CHAPTER

26 Gradient and Area under a Graph

CHAPTER

2

Mathematics SPM Chapter 26

© Penerbitan Pelangi Sdn. Bhd.

(c) 12

(u+30)(5)+(17–5)× 30 = 470

12

(5u+150)+360=470

12

(5u+150)=470–360

5u + 150 = 2 × 110 5u = 220 – 150

u = 705

= 14

3. (a) Durationoftime =105−80 = 25 minutes

(b) Thespeedofthevaninthefirst80minutes

= (180–60)km8060

h

= 120km43

h

=90kmh−1

(c) Thedistancetravelledinthelast45minutes = 60 – 0 =60km

4. (a) TheuniformspeedofA = 24 m s−1

(b) RateofchangeofspeedofA =GradientoflineRS

= 24 – 010 – 0

= 2.4 m s–2

(c) 3 12(8+18)× 244 – 3 1

2 (u+24)× 184 = 6

312 – 9u – 216 = 6 96 – 6 = 9u 90 = 9u

u = 90–––9

= 10

5. (a) Theuniformspeedoftheparticle = 8 m s–1

(b) Rateofchangeofspeedinthefirst3s

= 8 – 23 – 0

= 2 m s−2

(c) 12(2+8)(3)+8(5−3)+ 1

2(8+20)(t−5)=143

15 + 16 + 14(t−5)=143 31 + 14t−70=143 14t=143+70−31

t = 182––––14

= 13

6. (a) Length of time during which X moves withuniform speed

= 12 s

(b) RateofchangeofspeedofY =GradientoflineDE

= 30 – 012 – 0

= 2.5 m s−2

(c) DistancetravelledbyX=DistancetravelledbyY

12(12+18)× v = 1

2 × 12 × 30

15v = 180

v = 18015

= 12

7. (a) Theuniformspeedoftheparticle=20ms−1

(b) (i) (t−6)× 20 = 300

(t−6)= 300––––20

t−6=15 t = 15 + 6 = 21

(ii) Totaldistancetravelled

= 1—2

×(20+31)× 6 + 300 +

1—2

×(28−21)× 20

= 153 + 300 + 70 = 523 m

Average speed

= 523––––28

= 18.68 m s−1

3



Paper 2


= 10 – 0––––––6 – 0

= 5—3

m s−2

(b)Totaldistancetravelled=areaunderthegraph

= 1—2

×(16+10)× 10

= 130 m

2. (a) 1—2

× 10 × V = 80 5V = 80

V = 80–––5

= 16

(b) Distancetravelledatconstantspeed =(20−10)× 16 = 160 m

3. (a) 1—2

× (T+8)× 12 = 132

6(T+8)=132

T + 8 = 132––––6

T=22−8 = 14

(b) Averagespeed

= totaldistance––––––––––––totaltime

= 132––––14

= 9.429 m s−1

4. (a) Rateofchangeofspeedinthefirstt s = 2

Hence, 8 – 0–––––t – 0 = 2

8—t

= 2

t = 8—2

= 4

(b) Distancetravelledinthelast8s

= 1—2

×(8+12)×(14−6)

= 1—2(20)(8)

= 80 m

5. (a) v−4–––––2−0

= 2v−v––––––7 – 3

v−4–––––2

= v—4

4(v−4)=2v 4v−16=2v 4v−2v = 16 2v = 16 v = 8

(b) Distancetravelledinthefirst3s

= 1—2

× (4 ×8)×2+(3−2)× 8

= 12 + 8 = 20 m

6. (a) AveragespeedofparticleA

= 128 – 24––––––––16 – 0

= 104––––16

= 6.5 m s−1

(b) DistancetravelledbyparticleA =128−24 = 104 m

DistancetravelledbyparticleB

= 1—2

×(16+9)× 5

= 62.5 m

Hence,thedifferenceindistancebetweenA and B

=104−62.5 = 41.5 m

7. (a) Length of time that the particle moves withconstant speed

=65−20 = 45 s(b) Rateofchangeofspeedinthefirst20s

= 15 – 9––––––0 – 20

=−0.3ms−2

(c) Distancetravelledinthelast70s

=(65−20)× 9 + 1—2(9+15)(90−65)

= 405 + 300 = 705 m

4




= 10 – 5––––––3 – 0

= 5—3

m s−2

(b) Theuniformspeedoftheparticleis16ms−1.

(c) 1—2(5+10)(3)+ 1—

2(10+16)(T−3)=100.5

22.5 + 13(T−3)=100.5 22.5 + 13T−39=100.5 13T=100.5+39−22.5

T = 117––––13

= 9

9. (a) PeriodthatparticleAmoveswithconstantspeed =70−30 = 40 s

(b) RateofchangeofspeedofparticleAinthefirst30 s

= 6 – 0––––––30 – 0

= 0.2 m s−2

(c) DistancetravelledbyparticleA

= 1—2

×(90+40)× 6

= 390 m

DistancetravelledbyparticleB = 4 × 90 = 360 m

Differenceindistancebetweenthetwoparticles =390−360 = 30 m


=14−4 = 10 s

(b) Rateofchangeofspeedinthefirst4s

= 16 – 0––––––4 – 0

= 4 m s−2

(c) 1—2

× (T+10)× 16 = 248 8(T+10)=248 8T + 80 = 248 8T=248−80

T = 168––––8

= 21

11. (a) Rateofchangeofspeedfrom6thsecondto8thsecond

= 28 – 10–––––––8 – 6

= 18–––2

= 9 m s−2

(b) Distancetravelledinthelast5s

=(6−3)× 10 + 1—2(10+28)(8−6)

= 30 + 38 = 68 m

(c) Totaldistancetravelled

= 1—2(10+18)(3)+68

= 42 + 68 = 110 m

Average speed

= totaldistance

––––––––––––totaltime

= 110–––8

= 13.75 m s–1

12. (a) Averagespeedforthefirst8s

=

1—2(8)(24)

–––––––––8−0

= 96–––8

= 12 m s−1

(b) (i) 1—2

(t)(24)=156

12t = 156

t = 156––––12

= 13

(ii) 24–––––t – 8

= 2 24–––8

1–––––t – 8

= 1—4

t −8=4 t = 4 + 8 = 12

5



13. (a) Rateofchangeofspeedinthelast6s

= 50 – 20–––––––12 – 6

= 30–––6

= 5 m s−2

(b) Totaldistancetravelled

= 1—2(6+4)(20)+ 1—

2(20+50)(12−6)

= 100 + 210 = 310 m

(c) Averagespeed

= 310–––12

= 25.83 m s−1


=14−4 = 10 s

(b) (i) Totaldistancetravelled =12v + 8

1—2

×(18+10)× v = 12v + 8

14v = 12v + 8 14v−12v = 8 2v = 8 v = 4

(ii) Averagespeed

= 12v + 8–––––––18

= 12(4)+8–––––––––18

= 3.111 m s−1

15. (a) Lengthoftimethatthecyclistisatrest = from 1045 to 1110 = 25 minutes

(b) Averagespeed

= 2.8km––––––––2.5 hours

=11.2kmh−1

(c) Speedfrom1000to1045

= 12km––––––3—4

h

=16kmh−1

Speedfrom1045to1100 = 0

Speedfrom1110to1230

= (28−12)km––––––––––––––––

1 hour 20 minutes

= 16km––––––––––1 1—

3 hours

=12kmh−1

0

Time

28

1000 1230

2420161284

1045 1110

Speed (km h–1)

16. (a) DistancebetweenB and D =18−6 =12km

(b) (i) SpeedofthecarfromB to D

= 12km––––––––––––––––

(50−20)minutes

= 12km–––––––––30–––60

hour

=24kmh−1

(ii) Averagespeed

= 18km––––––––––

50 minutes

= 18km–––––––––50–––60

hour

=21.6kmh−1

Documents

26[A Math CD]