A particle is moving with a constant acceleration of 4 m/s^2 . Its speed at t=1 s is 4 m/s and t=3 s it is 12 m/s. What is the area under the position-time graph for the interval from t=1 s to t=3 s ?

Question

A particle is moving with a constant acceleration of 4 m/s^2 . Its speed at t=1 s is 4 m/s and t=3 s   it is 12 m/s. What is the area under the position-time graph for the interval from t=1 s to t=3 s ?

anonymous · Answer

$$v _{f}^2=v _{i}^2+2ax$$

Lets do t=1 first
Initial velocity is 0
Final velocity is 4
a is acceleration and that is 4 m/s^2
Plug in and find position of t=1s.

For t=2
Initial velocity would be 0 m/s
and final velocity would be 8 m/s due to it moving at an acceleration of 4 m/s^2

For t=3, do the same thing

Once you are done, graph it.
It should look this and you find the area of it.

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anonymous · Answer

I Followed your steps got the same graph and calculated the area it was 20 units
But i have choices to choose from and 20 isn't any of them
a) 8 m/s
b) 8 m
c) 12 m
d) 16 m
e) 16 m/s^2

anonymous · Answer

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anonymous · Answer

The area under the curve should be the distance traveled 
since position over time is velocity
Integral of velocity is position.
So it would be position or displacement from t=1 to t=3.
18m-2m=16m

I guess we start the area at the y=2 instead.
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anonymous · Answer

Not 100% sure..

anonymous · Answer

Its supposed to be a curve
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