Solve Distance from Velocity Time Graph - Vf2=Vo2+2ad

[ThomasMagnus]

The diagram is in the attachment.

This is my attempt at a solution.

Part 1 from t=1 to t=4
a=2
Vo=0
Vf=8

Vf²=Vo² +2ad

64=0+2(2)(D)

64= 4D

D=16m for the first part

Part 2 from t=5 to t=15
Vo=9
Vf=9
a=0

V=d/t
9=d/(15-5)
d=90m for the second part

there are also 8 blocks below that half-three quarter block between t8 and t9

Total distance= 90+16 +8=114m

The correct answer is 110m. Are they just rounding down? I think I am off because of that part between t=8 and t=9. I can't seem to final the distance in that interval. This is what I tried:
Vf=9
Vo=8
t=1
a=1

Vf²=Vo²+2ad
81=64+2D
17=2D
D=8.5m

However this does not seem to work...

Can you please help me with where I am going wrong?

Thanks!

 

[ThomasMagnus]

It seems to work if I use 2m/s^2 for the part between t=8 and t=9, but why would the acceleration be the same? The slope of the graph is changing in that part isn't it?

 

[2milehi]

Area under the curve is distance. With multiple guess, closest answer is usually correct.