Page 72 - eEISBN FINALIZE

iv. Distance during decelaration, S3 = the area under the graph. (area of triangle B-
C-17)).

1
= (6)(2.8)
3
2
= .
3

v. Let the total distance traveled, ST, and said during a uniform velocity, S2
Hence ; = + +
1
3
2

= 417
2
= 13(2.8)
2
= .
2

= 5.6 + 36.4 + 8.4

= .

or

= 0

1
= (13 + 23)(2.8)

2
= .

Example 3

A car moves from station A to station D in 40 seconds. During that travel, he went through the

station B and station C. Trip from station A to station B takes 10 seconds with an acceleration of
-2
16 ms . While traveling from station B to station C takes 15 seconds with an acceleration of 8 ms -
2 . For the last 15 seconds, it moves with uniform decelaration to stop at the station D. Calculate:

i. Maximum velocity is reached.

ii. Decelaration in the last 15 seconds.

iii. Total distance traveled.

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