2140 Words9 Pages

1. ht= -4.9t2+ 450, where t is the time elapsed in seconds and h is the height in metres.

a) Table of Values

t(s) | h(t) (m) | 0 | ht= -4.9(0)2+ 450= 450 | 1 | ht= -4.9(1)2+ 450= 445.1 | 2 | ht= -4.9(2)2+ 450= 430.4 | 3 | ht= -4.9(3)2+ 450= 405.9 | 4 | ht= -4.9(4)2+ 450=371.6 | 5 | ht= -4.9(5)2+ 450=327.5 | 6 | ht= -4.9(6)2+ 450= 273.6 | 7 | ht= -4.9(7)2+ 450= 209.9 | 8 | ht= -4.9(8)2+ 450= 136.4 | 9 | ht= -4.9(9)2+ 450=53.1 | 10 | ht= -4.9(10)2+ 450= -40 |

b) Average velocity for the first 2 seconds after the ball was dropped=

h2-h02-0 = 430.4-4502-0 = -19.62 = -9.8 m/s

c) Average velocity for the following time intervals:

i. 1≤t≤4 = h4-h14-1 = 371.6-445.14-1 = -73.53 = -24.5 m/s*…show more content…*

Substitute these values into the equation and simplify to find the instantaneous rate of change of a function.

b) Evaluate the following limits:

i. limx→0x-32x2-5 = 0-3202-5 = -3-5 = 35 = 0.6

ii. limx→22x2-7x+6x-2 = 222-72+62-2 = 00

Therefore, limx→22x2-7x+6x-2 = 2x-3x-2x-2 = limx→22x-3= 1

5. ht= -5t2+ 20t+1, where h is height in metres and t is the time in seconds.

a) h2=-522+ 202+1 = 21

h2+h= -52+h2+ 202+h+ 1 = -54+4h+h2+ 202+h+ 1 = -54+4h+h2+40+20h+1 = -20-20h-5h2+40+20h+1 = -5h2+21

Instantaneous velocity of the ball at t= 2 seconds is

lim h→0h2+h-h2h = -5h2+21-21h = -5h2h = h-5hh = -5h limh→0-5h= -50=0

Therefore the instantaneous velocity of the ball at t= 2 seconds is 0 m/s.

QUESTION 5 CONTINUED

b) dt= t2-8t+15, where d is measured in metres and t is the time in seconds.

The particles is at rest when t= 4 seconds and therefore the instantaneous velocity of the particle at t= 4 seconds should be 0 m/s.

d4=42- 84+15= -1 d4+h=4+h2-84+h+ 15 = 16+8h+h2-84+h+15 = 16+8h+h2-32-8h+15 = h2+16-32+15 = h2-1 Instantaneous velocity of the particle at t= 4 seconds is lim h→0d4+h-d4h = h2-1+1h = h2h = hhh = h limh→0h= 0

Therefore the instantaneous velocity of the particle at t= 4 seconds is 0 m/s and hence

a) Table of Values

t(s) | h(t) (m) | 0 | ht= -4.9(0)2+ 450= 450 | 1 | ht= -4.9(1)2+ 450= 445.1 | 2 | ht= -4.9(2)2+ 450= 430.4 | 3 | ht= -4.9(3)2+ 450= 405.9 | 4 | ht= -4.9(4)2+ 450=371.6 | 5 | ht= -4.9(5)2+ 450=327.5 | 6 | ht= -4.9(6)2+ 450= 273.6 | 7 | ht= -4.9(7)2+ 450= 209.9 | 8 | ht= -4.9(8)2+ 450= 136.4 | 9 | ht= -4.9(9)2+ 450=53.1 | 10 | ht= -4.9(10)2+ 450= -40 |

b) Average velocity for the first 2 seconds after the ball was dropped=

h2-h02-0 = 430.4-4502-0 = -19.62 = -9.8 m/s

c) Average velocity for the following time intervals:

i. 1≤t≤4 = h4-h14-1 = 371.6-445.14-1 = -73.53 = -24.5 m/s

Substitute these values into the equation and simplify to find the instantaneous rate of change of a function.

b) Evaluate the following limits:

i. limx→0x-32x2-5 = 0-3202-5 = -3-5 = 35 = 0.6

ii. limx→22x2-7x+6x-2 = 222-72+62-2 = 00

Therefore, limx→22x2-7x+6x-2 = 2x-3x-2x-2 = limx→22x-3= 1

5. ht= -5t2+ 20t+1, where h is height in metres and t is the time in seconds.

a) h2=-522+ 202+1 = 21

h2+h= -52+h2+ 202+h+ 1 = -54+4h+h2+ 202+h+ 1 = -54+4h+h2+40+20h+1 = -20-20h-5h2+40+20h+1 = -5h2+21

Instantaneous velocity of the ball at t= 2 seconds is

lim h→0h2+h-h2h = -5h2+21-21h = -5h2h = h-5hh = -5h limh→0-5h= -50=0

Therefore the instantaneous velocity of the ball at t= 2 seconds is 0 m/s.

QUESTION 5 CONTINUED

b) dt= t2-8t+15, where d is measured in metres and t is the time in seconds.

The particles is at rest when t= 4 seconds and therefore the instantaneous velocity of the particle at t= 4 seconds should be 0 m/s.

d4=42- 84+15= -1 d4+h=4+h2-84+h+ 15 = 16+8h+h2-84+h+15 = 16+8h+h2-32-8h+15 = h2+16-32+15 = h2-1 Instantaneous velocity of the particle at t= 4 seconds is lim h→0d4+h-d4h = h2-1+1h = h2h = hhh = h limh→0h= 0

Therefore the instantaneous velocity of the particle at t= 4 seconds is 0 m/s and hence

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