Page 69 - eEISBN FINALIZE

During acceleration: = 0, = 9.8 −1 , = 0.6 −2

2
= + 2
2
1
2
2
(9.8) = 0 + 2(0.6)
1
(9.8) 2
=
1
2(0.6)
= .
1

During decelaration: = 9.8 −1 , = 0, = 20

1
= ( + )
3
2
1
= (9.8 + 0)20
3
2
=
3

Hence,

= + +

3
2
1
= 80.03 + 2700 + 98

=

ii. Suppose:The amount of time taken, tT.

During time acceleration, s1.

The current time uniform velocity, s2

The current time decelaration, S3 = 20 s.

= + +
1
2

3

During acceleration: = +
1
9.8 = 0 + (0.6)
1
9.8
=
1
0.6
= .
1

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