Cambridge Additional Mathematics

Answers 495

e

11 Hint: Find

dy

dx

, then determine the nature of the stationary

points.

12 Hint: Show that as x! 0 ,f(x)!¡1,

and asx!1,f(x)! 0.

13 aHint: Findf^0 (x), then determine the nature of the

stationary points.

bHint: Show thatf(x)> 1 for allx> 0.

EXERCISE 14C.1

1a 7 ms¡^1 b(h+5)ms¡^1

c 5 ms¡^1 =s^0 (1)is the instantaneous velocity att=1s

daverage velocity=(2t+h+3)ms¡^1 ,

lim

h! 0

(2t+h+3)=2t+3ms¡^1 is the instantaneous

velocity at timetseconds.

2a¡ 14 cm s¡^1 b(¡ 8 ¡ 2 h)cm s¡^1

c¡ 8 cm s¡^1 =s^0 (2)

) instantaneous velocity=¡ 8 cm s¡^1 att=2

d¡ 4 t=s^0 (t)=v(t) is the instantaneous velocity at time

tseconds.

3a^23 cm s¡^2 b

2

p

1+h¡ 2

h cm s

¡ 2

c 1 cm s¡^2 =v^0 (1)is the instantaneous accn. at t=1s

d

1

p

t

cm s¡^2 =v^0 (t), the instantaneous accn. at timet

4avelocity at t=4 bacceleration at t=4

EXERCISE 14C.2

1av(t)=2t¡ 4 cm s¡^1 , a(t)=2cm s¡^2

bs(0) = 3cm, v(0) =¡ 4 cm s¡^1 , a(0) = 2cm s¡^2

The object is initially 3 cm to the right of the origin and is

moving to the left at 4 cm s¡^1. It is accelerating at 2 cm s¡^2

to the right.

cs(2) =¡ 1 cm, v(2) = 0cm s¡^1 , a(2) = 2cm s¡^2

The object is instantaneously stationary, 1 cm to the left of

the origin and is accelerating to the right at 2 cm s¡^2.

dAt t=2, s(2) = 1cm to the left of the origin.

ef 06 t 62

2av(t)=98¡ 9 : 8 tms¡^1 , a(t)=¡ 9 : 8 ms¡^2

bs(0) = 0m above the ground, v(0) = 98ms¡^1 skyward

ct=5s Stone is 367 : 5 m above the ground and moving

skyward at 49 ms¡^1. Its speed is decreasing.

t=12s Stone is 470 : 4 m above the ground and moving

groundward at 19 : 6 ms¡^1. Its speed is increasing.

d 490 m e 20 seconds

3a 1 : 2 m

bs^0 (t)=28: 1 ¡ 9 : 8 trepresents the instantaneous velocity of

the ball.

ct=2: 87 s. The ball has reached its maximum height and is

instantaneously at rest.

d 41 : 5 m

ei 28 : 1 ms¡^1 ii 8 : 5 ms¡^1 iii 20 : 9 ms¡^1

s^0 (t)> 0 when the ball is travelling upwards.

s^0 (t) 60 when the ball is travelling downwards.

f 5 : 78 s

gs^00 (t)is the rate of change of s^0 (t), or the instantaneous

acceleration.

4av(t)=3t^2 ¡ 18 t+24ms¡^1 a(t)=6t¡ 18 ms¡^2

bx(2) = 20, x(4) = 16

ci 06 t 62 and 36 t 64 ii 06 t 63

d 28 m

5av(t) = 100¡ 40 e

¡t 5

cm s¡^1 , a(t)=8e

¡ 5 t

cm s¡^2

bs(0) = 200cm on positive side of origin

v(0) = 60cm s¡^1 , a(0) = 8cm s¡^2

cdafter 3 : 47 s

EXERCISE 14D

1a$118 000 b

dP

dt

=4t¡ 12 ,$ 1000 s per year

c dP

dt

is the rate of change in profit with time

di 06 t 63 years iit> 3 years

eminimum profit is $100 000whent=3

-1 0 3 s

24 t

++-

0

v(t)

0 16 20 x

t=4

t=2

1 3 t

++-

0

s(t)

2 t

• 
  
  
  
  • 0
  
  
  
  

v(t)

t

+

0

a(t)

t

+

0 20

s(t)

(^10) t

• 
  
  
  
  • 0 20
    v(t)
    t
  
  
  
  

• 0 20

a(t)

t( )s

v(t) (cm s-1)

60

v(t) = 100¡¡

O

6ax(0) =¡ 1 cm, v(0) = 0cm s¡^1 , a(0) = 2cm s¡^2

bAtt=¼ 4 seconds, the particle is (

p

2 ¡1)cm left of the

origin, moving right at

p

2 cm s¡^1 , with increasing speed.

cchanges direction whent=¼,x(¼)=3cm

d 06 t 6 ¼ 2 and ¼ 6 t 632 ¼

7 Hint: Assume that s(t)=at^2 +bt+c

s^0 (t)=v(t)ands^00 (t)=a(t)=g

Show thata=^12 g,b=v(0),c=0.

8a 0 : 675 s

biS^0 (t)=u+atms¡^1 iit=¡

u

a

s

iii a=¡^64099 ¼¡ 6 : 46 ms¡^2

iv Hint: Substitutet=¡

u

a

into S(t).

v If the speeduis doubled, then the braking distance is

quadrupled ( 22 =4times).

3 t

• 
  
  
  
  • 0
  
  
  
  

a(t)

(2¼,¡2)

¼ 2 ¼

2

-2

x

y

local min.

()^562 ¼,-3~`3 stationary inflection

local max.

()¼ 62 ,^33 ~`

y =¡¡¡¡¡sin(2x) + 2cosx

O

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