20[anal add math cd]

7/24/2019 20[Anal Add Math CD]

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Additional Mathematics SPM Chapter 20

Penerbitan Pelangi Sdn. Bhd.

1. (a) Distance OA= 5 m

Distance OB= 2 m

(b) Displacement ofAfrom O= 5 m

Displacement ofBfrom O= 2 m

2. (a) Displacement after 5 seconds

= 52

+ 3(5) = 40 m

(b) Distance travelled in the rst 4 seconds

= s4 s

0

= [42+ 3(4)] (0)

= 28 m

(c) s5 s

3

= [52+ 3(5)] [32+ 3(3)]

= 22 m

(d) Distance travelled during the 3rdsecond

= s3 s

2

= [32+ 3(3)] [22+ 3(2)]

= 8 m

3. (a) s= t2 4t

When s= 0,

t2 4t= 0

t(t 4) = 0

t= 0, 4

After 4 seconds, the particle returns to O.

(b) When s= 4,

4 = t2 4t

t2

4t+ 4 = 0 (t 2)2= 0

t= 2

After 2 seconds, the particle is at 4 m on the left

of O.

(c) s0

t2 4t0

t(t 4) 0

s

t

0 4

0 t4 is the time when the particle is on the

left of O.

(d) s0

Based on diagram in (c),

t4 is the time when the particle is on the right

of O.

4. (a) s= t(t 4)

= t2 4t

t 0 2 4 5

s 0 4 0 5

s

t0 542

4

5

(b) (i) The furthest distance on the left is 4 m from

O.

(ii) Total distance travelled in the rst 5 seconds

= 2 4 + 5

= 13 m

(iii) The displacement after 5 seconds is 5 m on

the right of O.

CHAPTER

20Motion along aStraight Line


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(iv) The time when the particle passes through

Oagain is 4 s.

(v) t 4 is the time interval when the particle

is on the right-hand side of O.

5. (a) s= t2 9

OAt= 0

When t= 0, s= 9

OA= 9 m

(b) s= 0

t2 9 = 0

(t+ 3)(t 3) = 0

t= 3, t= 3 is ignored

(c) s0

t2 9 0

(t+ 3)(t 3) 0

s

t0

33

0 t3 is the time interval when the particle

is on the left-hand side of O.

(d) s= 42 9

= 7 m

(e) s4 s

3

= (42 9) (32 9)

= 7 m

(f) s= t2 9

s

t0

7

9

3 4

6. (a) v= 2t+ 6

Initial velocity = 2(0) + 6

= 6 m s1

(b) When v= 10,

10 = 2t+ 6

2t= 4

t= 2

After 2 seconds passing through O, its velocity

is 10 m s1.

(c) When t= 1,

v= 2(1) + 6

= 8

Therefore, the particle is moving towards right.

7. (a) s= 3t2 12t

v=ds

dt = 6t 12

(b) Initial velocity = 6(0) 12

= 12 m s1

Its direction of motion is towards left.

(c) When v= 0,

6t 12 = 0

t= 2

Substitute t= 2 into s= 3t2 12t,

s= 3(2)2 12(2)

= 12 m

When the particle is instantaneous at rest, the time

is 2 s and the displacement is 12 m on the left of

O.

(d) v0

6t 120

t2

The time interval is 0 t2 when the particle

is moving towards left.

8. (a) v= t(t 4)

t 0 2 4v 0 4 0

v

t0

4

42

(b) (i) Minimum velocity is 4 m s1.

(ii) The time interval when the particle moves

towards left is 0 t4.

(iii)4

0 v dt =

4

0 (t2 4t) dt

= t3

3

2t24

0

=43

3

2(4)2 (0)

= 1023

m

Therefore, the distance travelled in the rst

4 seconds is 1023

m.


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9. (a) v= 3t2+ 8t

s= v dt

= (3t2+ 8t) dt = t3+ 4t2+ c

When t= 0, s= 0 c= 0

s= t3+ 4t2

(b) When v= 0,

3t2+ 8t = 0

t(3t+ 8) = 0

t= 0,83

The time when the particle stops instantaneously

for the second time is83

s.

(c) v0

3t2+ 8t0

t(3t+ 8) 0

v

t0

8

3

Time interval when the particle travelled in the

right direction is 0 t8

3

.

(d) Distance travelled = 83

0

v dt

= 83

0

(3t2+ 8t) dt

= t3+ 4t283

0

= 83

3

+ 4 83

2

(0) = 9

1327

m

(e) The maximum distance of the particle is whenv= 0, that is, t=

83

.

Maximum distance = t3+ 4t2

= 83

3

+ 4 83

2

= 913

27

m

(f) v= 3t2+ 8t

t 043

83

3

v 0 513

0 3

v

t0

3

8

3

4

3

1

3

3

5

10. (a) v= t3 4t

When v= 0,

t3 4t = 0

t(t2 4) = 0

t = 0, t2= 4

t = 2, t= 2 is ignored

After 2 seconds, the particle stops instantaneously.

(b) The distance travelled in the rst 3 seconds

= 2

0 v dt+

3

2 v dt

2

0 v dt=

2

0 (t3 4t) dt

= t4

4

2t22

0

= 24

4

2(2)2 (0) = 4

3

2 v dt= t

4

4

2t23

2

= 34

4

2(3)2 24

4

2(2)2 = 6

14

m

The distance travelled in the rst 3 seconds

= 4 + 614

= 1014 m

(c) v0

t3 4t0

t(t2 4) 0

From the graph,

0 t2 is the time interval when the particle

is moving towards left.

(d) When t= 5,

Velocity, v= 53 4(5)

= 105 m s1


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11. (a) s= t3 2t2+ t

Velocity, v=ds

dt

= 3t2 4t+ 1

(b) Acceleration, a=dv

dt

= 6t 4

(c) When v= 0,

3t2 4t+ 1 = 0

(3t 1)(t 1) = 0

t=13

, 1

The acceleration of the particle when it stops at

second time

= 6(1) 4

= 2 m s2

(d) When a= 8, 6t 4 = 8

t= 2

The displacement when a= 8

= 23 2(2)2+ 2

= 2 m on the right of O

(e) Minimum velocity is when a= 0.

6t 4 = 0

t=46

=23

Minimum velocity

= 3t2 4t+ 1

= 3 23

2

4 23+ 1

= 13

m s1

(f) s= t3 2t2+ t

t 0 1 2

s 0 0 2

s

t0

21

2

(g) v= 3t2 4t+ 1

= (3t 1)(t 1)

t 013

1 2

v 1 0 0 5

v

t0 211

3

5

1

(h) a= 6t 4

t 023

2

a

4 0 8

a

t0 22

3

8

4

(i) Distance travelled in the rst second

= 13

0

v dt+ 1

13

v dt

v

t0

1

11

3

13

0

v dt= 13

0

(3t2 4t+ 1) dt

= t3 2t2+ t13

0

= 133 2 13

2+ 1

3 (0) =

427

m

1

13

v dt= t3 2t2+ t1

13

= [13 2(1)2+ 1] 13

3

2 13

2

+13

= 0 427

=

427

m


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Distance travelled in the rst second

=4

27

+4

27

=8

27

m

12. (a) a= 6t

v= a dt

= 6t dt = 3t2+ c

Given t= 0, v= 12,

c= 12

v= 3t2+ 12

(b) s= v dt

= (3t2+ 12) dt

= t3+ 12t+ c

Given t= 0, s= 0,

c= 0

s= t3+ 12t

(c) (i) Displacement = 13+ 12(1)

= 13 m on the right of O

(ii) Velocity = 3(1)2+ 12

= 15 m s1

(iii) Acceleration = 6t

= 6(1) = 6 m s2

(d) When v= 24,

3t2+ 12 = 24

3t2= 12

t2= 4


Acceleration when v= 24

= 6 2

= 12 m s2

(e) (i) s= t3+ 12t

Distance travelled after 2 seconds = 23+ 12(2)

= 32 m

(ii) s2 = 32 m

s1 = 13+ 12(1)

= 13 m

Distance travelled during the second second

= s2 s

1

= 32 13

= 19 m

(iii) s= t3+ 12t

= 43+ 12(4)

= 112 m

13. (a) v= 6t 10

a=dv

dt

a= 6

(b) s= v dt

s= (6t 10) dt s= 3t2 10t+ c

Given t= 0, s= 0, c= 0

s= 3t2 10t

(c) s0

3t2 10t0

t(3t 10) 0

s

t0

10

3

Therefore, the time interval for the particle on the

left of Ois 0 t10

3

.

(d) v0

6t 10 0

t10

6

t53

Therefore, the time interval when the particle is

moving towards direction of left is 0 t53

.

(e) v= 0 when sis the maximum

6t 10 = 0

t= 53

Maximum displacement on the left

= 3 53

2

10 53

=25

3

50

3

= 25

3

m


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(f) Distance travelled from t= 3 to t= 5

=5

3 v dt

=5

3 (6t 10) dt

= 3t2 10t5

3

= [3(5)2 10(5)] [3(3)2 10(3)] = 25 (3)

= 28 m

(g) (i) v= 6t 10

v

t

B

A

0 4

(4, 14)

5

310

Total distance travelled in the rst 4 seconds

= 12

10 53+ 1

214

73

=25

3

+49

3

=74

3

Average speed =

74

3

4

= 616

m s1

(ii) Average velocity =Total displacement

Time

=84

= 2 m s1

1. (a) Given a= 2t 8

For minimum velocity, a= 0

2t 8 = 0

t= 4

From a= 2t 8,

v= (2t 8) dt v= t2 8t+ c

Given t= 0, v= 12

c= 12

v= t2 8t+ 12

Minimum velocity, v= 42 8(4) + 12

= 4 m s1

(b) v0

t2 8t+ 12 0

(t 2)(t 6) 0

v

t0

2 6

2 t6

(c)v

t0

2 6

4

12


= Area of shaded region

=2

0 (t2 8t+ 12) dt+

4

2 (t2 8t+ 12) dt

= t3

3 4t2+ 12t

2

0

+ t3

3 4t2 12t

4

2

= 23

3 4(2)2+ 12(2)+ 4

3

3 4(4)2+ 12(4)

23

3 4(2)2 12(2)

= 102

3+ 5

1

3

= 16 m

2. (a) v= t2+ 3t+ 4

At the pointR, v = 0

t2+ 3t+ 4 = 0

t2 3t 4 = 0

(t 4)(t+ 1) = 0

t= 4

After 4 seconds, the particle travels from point O

to pointR.

(b) a=dv

dt

= 2t+ 3

The acceleration at pointR

= 2(4) + 3

= 5 m s2


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(c) For maximum velocity,

a= 0

2t+ 3 = 0

t=32

Maximum velocity

= 322

+ 3 32+ 4 = 6

14

m s1

3. (a) v= t2+ 8t 7

(i) Initial velocity

= (0)2+ 8(0) 7

= 7 m s1

(ii) v0

t2+ 8t 7 0

(t+ 1)(t 7) 0

v

t0

71

7

Therefore, the time interval during which the

particle moves towards right is 1t7.

(iii) a0

2t+ 8 0

2t8

t4

Therefore, the time interval during which

the acceleration is negative is t4.

(b) v= t2+ 8t 7

= (t+ 7)(t 1)

t 0 1 7

v 7 0 0

v

t0

71

7

(c) The total distance travelled during the first

7 seconds

= 1

0 v dt+

7

1 v dt

1

0 v dt= (t2+ 8t 7) dt

= t3

3

+ 4t2 7t1

0

= 13

+ 4 7 (0)

= 313

m

7

1 v dt= t

3

3

+ 4t2 7t7

1

= 73

3

+ 4(7)2 7(7) 13

+ 4 7

= 36 m

Total distance travelled = 313

+ 36

= 3913

m

4. (a) vQ

= 2t2 4t 16

dv

Q

dt= 4t 4

vQis minimum when

dvQ

dt

= 0,

4t 4 = 0

t= 1

Therefore, the minimum velocity of Q

= 2(1)2 4(1) 16

= 18 m s1

(b) Particle Qstops at C,

vQ= 0

2t2 4t 16 = 0

t2 2t 8 = 0

(t 4)(t+ 2) = 0

t= 4, t= 2 is ignored v

Q = 2t2 4t 16

sQ = (2t2 4t 16) dt

=23

t3 2t2 16t+ c

Given t= 0, SQ= 0, c= 0

sQ=

23

t3 2t2 16t


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When t= 4,

BC=23

(4)3 2(4)2 16(4)

= 5313

m

AC= 60 5313

= 623

m

(c) When Qat C, t= 4

Displacement of P= 4 3

= 12 m

Difference in distance between Pand Q

= 12 623

= 513

m

1. (a) s= 3t2 12t

Displacement = s1 s

0

= [3(1)2 12(1)] (0)

= 9 m

The distance travelled in the rst second is 9 m.

(b) s0

3t2 12t0

3t(t 4)

0

s

t0

4

Therefore, the time interval for the particle to be

on the left of Ois 0 t4.

(c) When the particle passes through Oagain, s= 0

3t2 12t= 0 3t(t 4) = 0


v=ds

dt

= 6t 12

The velocity when the particle passes through O

again

= 6(4) 12

= 12 m s1

(d) Maximum distance is when v= 0,

6t 12 = 0

t= 2

Displacement from O= 3(2)2 12(2)

= 12 m

Therefore, the maximum distance from O is

12 m.

(e) When s= 15,

3t2 12t= 15

3t2 12t 15 = 0

t2 4t 5 = 0

(t+ 1)(t 5) = 0


The velocity = 6(5) 12

= 18 m s1

2. (a)

BCA

sP= t2+ 4t, sQ= t

2 8t

P

60 m

Q

sP= t2+ 4t, s

Q= t2 8t

Assume Pand Q meet at Cafter kseconds.

Distance travelled by P+ Distance travelled by

Q= 60

(k2+ 4k) + (8k k2) = 60

12k= 60

k= 5

Substitute t= 5 into sP = t2+ 4t,

sp = 52+ 4(5) = 45 m

The two particles meet after 5 seconds they pass

throughAandBrespectively and it is at 45 m on

the right ofA.

(b) When Qreverses, vQ= 0

2t 8 = 0

t= 4

Substitute t= 4 into sQ= t2 8t,

sQ = 42 8(4)

= 16 m

After 4 seconds particle Qpasses through B, itreverses its direction at a distance of 16 m on the

left ofB.

(c) Velocity of object P, vP

= 2t+ 4

When t= 4,

vP

= 2(4) + 4

= 12 m s1


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(d) sQ= t2 8t

t 0 8 10

s 0 0 20

s

t0

108

20

(e) vQ0

2t 8 0

t4

0 t4 is the time interval for object Qmoves

in the direction of left.

3. (a) v= t(4 3t)

v= 4t 3t2

When the particle reverses, v= 0

4t 3t2= 0

t(4 3t) = 0

t=43

, t= 0 is ignored

Given v= 4t 3t2,

a= 4 6t

At t= 43

,

a= 4 6 43

= 4 m s2

(b) s= (4t 3t2) dt = 2t2 t3+ c

Given t= 0, s= 0, c= 0

s= 2t2 t3

When t= 1, s= 2(1)2 (1)3

= 1 m Distance travelled during the first second is

1 m.

(c) v= t(4 3t)

t 043

2

v 0 0 4

v

t0

24

3

4

Distance travelled during the rst 2 seconds

= 43

0

v dt+ 2

43

v dt

43

0

v dt= 43

0

(4t 3t2) dt

= 2t2 t343

0

= 2 43

2

43

3

(0) = 1 5

27m

2

43

v dt = 2t2 t32

43

= [2(2)2 23] 2 43

2

43

3

= 1

527

m


= 15

27

2

= 210

27 m

(d) The average velocity in the rst 2 seconds

=Displacement

Time

When t= 2, s= 2(2)2 23

= 0

Therefore, the average velocity is 0 m s1.

4. (a) v= t2 6t k

Given v= 0, when t= 3

0 = 32 6(3) k

k= 9

(b) v= t2 6t+ 9

When t= 0, v= 9 m s1

The velocity at Ois 9 m s1.


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(c) Distance travelled during the third second

=3

2 v dt

=3

2 (t2 6t+ 9) dt

= t3

3

3t2+ 9t3

2

= 33

3

3(3)2+ 9(3) 23

3

3(2)2+ 9(2) =

13

m

5. (a) (i) a= 2t 3

v = a dt

= (2t 3) dt = t2 3t+ c

Given t= 0, v= 10, c= 10 v= t2 3t 10

For minimum velocity, a= 0

2t 3 = 0

t=32

Minimum velocity = t2 3t 10

= 32

2

3 32 10

=94

92

10

= 12

1

4 m s

1

(ii) a0

2t 3 0

t32

The range of time for particle to decelerate

is 0 t32

.

(b) (i) v= t2 3t 10

= (t 5)(t+ 2)

t 032 5 6

v 10 1214

0 8

, 12

v

t0

3

2

1

4

65

10

8

(ii) Distance from t= 0 to t= 5

=5

0 v dt

=5

0 (t2 3t 10) dt

=

t3

3

3

2

t2 10t

5

0

= 53

3

32

(5)2 10(5) (0)

= 4556

m

Distance from t= 5 to t= 6

=6

5 v dt

= t3

3

32

t2 10t6

5

=

63

3

32

(6)2 10(6)

53

3

32

(5)2 10(5) = 3

56

m


+ 356

= 4923

m

6. (a) a= 10 m s2

v= 10 dt v= 10t+ c

Given t= 2, v= 15, c= 35

v= 10t 35

(i) At point O, t= 0,

Its velocity = 10(0) 35

= 35 m s1

(ii) s= (10t 35) dt s= 5t2 35t+ c

Given t= 0, s= 0,

c= 0 s= 5t2 35t

When v= 25,

10t 35 = 25

t = 6

The displacement of the object

= 5(6)2 35(6)

= 30 m

= 30 m on the left of O


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(b) When the object reverses its direction, v= 0

10t 35 = 0

t=35

10

t=72

s= 5t2

35t

= 5 72

2

35 72

= 6114

m

The distance travelled is 6114

m.

(c) At O, s= 0

5t2 35t= 0

5t(t 7) = 0

t= 0, 7

Velocity when the object at Oagain = 10(7) 35

= 35 m s1

1. (a) For sA= t t2

When t= 4, sA= 0

0 = 4 16

= 4

(b) sA = 4t t2...................................

sB = t............................................

WhenAandBmeet,

sA= s

B

4tt2=t

t2 4t+ t= 0

t2 3t= 0

t(t 3) = 0

t= 3, t = 0 is ignored as it is the starting

time

(c) sA = 4t t2

vA= 4 2t

For sAto be maximum, v

A= 0

4 2t= 0

t= 2

Maximum displacement ofA

= 4(2) 22

= 4 m

2. (a) s= t3+ pt2+ qt

Given s= 0, t= 3

0 = 33+ p(3)2+ q(3)

9p+ 3q+ 27 = 0

3p+ q+ 9 = 0..........................

s= t3+ pt2+ qt

Given s= 6 when t = 1

6 = 1 +p(1) + q(1)

p+ q+ 7 = 0 ...............................

, 2p+ 2 = 0 p= 1

Substitutep= 1 into , 1 + q+ 7 = 0

q= 6

Therefore,p= 1, q= 6

(b) s= t3 t2 6t

v= 3t2 2t 6

a= 6t 2

When a= 0,

6t 2 = 0

t=13

Velocity = 3 132

2 13 6

=13

23

6

= 19

3 = 6

13

m s1

3. (a) (i) v= pt qt2

When t= 4, v= 0

0 = 4p 16q

p= 4q..................................

a=p 2qt

Given a= 4 when t= 4

4 =p 2q(4)

4 =p 8q........................

Substituteinto , 4 = 4q 8q

4 = 4q

q= 1

Substitute q = 1 into , p= 4

Therefore,p= 4, q= 1


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(ii) v= 4t t2

s= 2t213

t3+ c

When t= 0, s= 0

c= 0

s= 2t213

t3

When at Oagain, s= 0

2t213

t3= 0

t22 13

t= 0 2

13


t= 6

(iii)

0 4 6

v

t

Distance travelled from t = 0 to t= 4

= 2(4)2 13

(4)3 0 = 10

23

m

Distance travelled from t= 4 to t= 6

= 2(6)2 13

(6)3 2(4)2 13

(4)3

= 0 1023

= 1023

= 1023

m

Total distance travelled in the first

6 seconds

= 2 1023

= 2

32

3

=64

3

= 2113

m

(b) s= 2t213

t3

When t= 0, s= 0

When s= 0,

2t213

t3= 0

t22 13 t= 0 2

13

t= 0

t= 6

For maximum s, v= 0

v= 4tt2

0 = 4t t2

t(4 t) = 0

t= 4

When t= 4,

s= 2(4)2

1

3 (4)3

=32

3

0

4 6

32

3

s

t

4.

(a) When t= 0, vB = 0

2

4 = 4 cm s1

(b) vA= 0 forAturning back

4 + 6t 4t2= 0

4t2 6t 4 = 0

2t2 3t 2 = 0

(2t+ 1)(t 2) = 0

t 2 = 0

t= 2 s, t= 12

is ignored

(c) vA = 4 + 6t 4t2

sA

= 4t+ 3t243

t3+ c

When t = 0, sA= 0

c= 0

sA= 4t+ 3t2

43

t3

When t = 2,

sA= 4(2) + 3(2)2

43

(2)3

= 913

cm


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Distance travelled from t= 2 to t= 3,

3

2 v dt= 4t+ 3t2 43 t

33

2

= 4(3) + 3(3)2 43

(3)3 9 13

= 61

3

cm


+ 613

= 1523

cm

(d) vA= 4 + 6t 4t2

a= 6 8t

For maximum vA, a= 0

6 8t= 0

t=68

t=

3

4 s

(e) vB= t2 4

When vB= 0,

t2 4 = 0

t= 2

sB = (t2 4) dt

=13

t3 4t+ c

When sB= 10, t= 0

c= 10

sB= 1

3t3 4t+ 10

When t= 2,

sB=

13

(2)3 4(2) + 10

= 423

cm

When t= 2,

sA= 9

13

cm

The distance betweenAandB

= 913 4

23

= 423

cm

5. vA= 30 4t

aB= 10 + 6t

(a) Initial velocity ofA

= 30 4(0)

= 30 m s1

(b) aA= a

B

4 = 10 + 6t

6t= 6

t= 1 s

(c) sA = (30 4t) dt

= 30t 2t2+ c

When t= 0, sA= 0

c= 0

sA = 30t 2t2

aB= 10 + 6t

vB= 10t+ 3t2+ c

When t= 0, vB= 20

c= 20

vB= 10t+ 3t2+ 20

sB= 5t2+ t3+ 20t+ d

When t= 0,sB= 0 d= 0

sB= 5t2+ t3+ 20t

WhenAandBmeet

sA= s

B

30t 2t2= 5t2+ t3+ 20t

t3 3t2 10t= 0

t(t2 3t 10) = 0

t(t 5)(t+ 2) = 0

t= 5 s, t= 0 and t= 2 are ignored

(d) WhenAat rest, vA= 0 30 4t= 0

t=30

4

t=15

2

sA

= 30 152 2 15

2

2

= 112.5 m

6. (a) v= 6 2t

v

0 6 2t0

2t6

t3

Therefore, 0 t3 is the range of time during

which the particle is moving towards B.


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14



(b) s = (6 2t) dt = 6t t2+ c

When t= 0, s= 0

c= 0

s= 6t t2

10 = 6t t2

t2 6t+ 10 = 0

b2 4ac= ( 6)2 4(1)(10)

= 36 40

= 4 0

It shows that the particle never reachesB.

(c) When s= 0,

6t t2= 0

t(6 t) = 0

t= 6 s

Therefore, t= 6 s is the time when the particle is

atAagain.

(d) When v= 0,

6 2t= 0

t= 3

This is time when particle turning back

When t= 5,

v= 6 2(5)

= 4

03 5

6

4

v

t


=12

3 6 +12

2 4

= 13 m

(e) s= 6t t2

t 0 3 6

s 0 9 0

0 3 6

9

sp

t

7. (a) a= k

v= kdt = kt+ c

Whent= 1, v= 8

8 = k+ c...............................

Whent= 3, v= 18 18 = 3k+ c...........................

, 10 = 2k

k= 5

(b) From (a), substitutek= 5 into , 8 = 5 + c

c= 3

v= 5t+ 3

8. (a) a= 10

v= 10t+ c

When t= 0, v= 30

c= 30

v= 10t+ 30

(b) For maximum height, v= 0

10t+ 30 = 0

t= 3

s= (10t+ 30) dt s= 5t2+ 30t+ c

When t= 0, s= 35

c= 35

s= 5t2+ 30t+ 35

Maximum height = 5(3)2+ 30(3) + 35

= 45 + 90 + 35

= 80 m

(c) When s= 0,

5t2+ 30t+ 35 = 0

t2 6t 7 = 0

(t 7)(t+ 1) = 0


Substitute t= 7 into v= 10t+ 30, v= 10(7) + 30

= 40 m s1

Therefore, the velocity just before it hits the

ground is 40 m s1.


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9. (a) h= kt2+ pt

v= 2kt+ p

When t= 0, v= 10

10 =p

v= 2kt+ 10

a= 2k

Given a= 8 8 = 2k

k= 4

Therefore,p= 10, k= 4

(b) v= 2(4)t+ 10

v= 8t + 10

For maximum height, v= 0

8t+ 10 = 0

t=10

8

=54 s

(c) h= 4t2+ 10t

At the surface of the sea,

h= qwhen t= 4

q= 4(4)2+ 10(4)

= 64 + 40

= 24

q= 24

20[anal add math cd]

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