I'm taking physics and trying to figure out the 4 questions to this problem.

Explanation:

At t = 1 velocity = 0

At t = 3 velocity = 3

slope of the line = 3-0 / 3-1 = 3/2

At t = 2

velocity = 3/2 x ( 2 - 1 )

= 1.5 m /s

velocity at t = 2 is 1.5 m /s

Position at t = 2 :

displacement at t = 2

area of graph upto  t = 2

= 1 / 2 x 1 x 1.5 = .75

position at t = 2 :

= initial position + displacement

= 10 + .75 = 10.75 m

position at 6 s :

displacement at t = 6

area of curve upto  t = 6

= 1 / 2 x 2 x 3 +  3 x 3 + 1/2 x 3 x ( 4.5 - 3 )

= 3 + 9 + 2.25

= 14.25 m

position at t = 6

= initial position + displacement

= 10 + 14.25 = 24.25 m

position at 9 s :

displacement at t = 9

area of curve upto  t = 9

= 1 / 2 x 2 x 3 +  4 x 3 + 1/2 x 4 x ( 5 - 3 )- 1/2 x 2 x 1.5

= 3 + 12 + 4 - 1.5

= 17.5  m

position at t = 9

= initial position + displacement

= 10 + 17.5 = 27.5  m

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