Page 70 - eEISBN FINALIZE

During uniform velocity: = = 9.8 −1 , = 2700

1
= ( + )
2 2
1
2700 = (9.8 + 9.8)
2 2
2700
=
2
9.8
= .
2

Hence: = + +

1
3
2
= 16.33 + 275.51 + 20

= .

Example 2

-1
A car starts from rest and moves with constant acceleration until it reaches a speed of 2.8 ms in
4 seconds. It maintains this speed for 13 seconds before stopping with uniform decelaration for 6

seconds.

Draw a velocity-time graph and find:

i. The acceleration of the car early.

ii. Train delays.

iii. The distance traveled during acceleration.

iv. Distance traveled in the decelaration.

iv. Total distance traveled.

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