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An object moves with constant acceleration 4.00 m/s^2 and over a time interval reaches a final velocity of 12.0 m/s. a) If its initial velocity is -6.00 m/s, what is its displacement during the time interval. b) What is the total distance it travels during the time interval in part b?

Physics
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specify what part b of the motion is?
oops, it's supposed to be "What is the total distance it travels during the time interval in part a" not part b. Typo.

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Other answers:

27/2 and total distance equals 45/2
Uh, how do you figure? I mean, what formula did you use? (kinematics)
\[v ^{2}-u ^{2} = 2aS\] v=12,u=-6 and a=4 u get total displacement S..
for, total distance use the same kinematic equation but twice once till the velocity becomes zero thats from u=-6 to v=0; then again from u=0 to v=12
I'm guessing u=initial velocity and v=final velocity. I get v, but why u?
And why stop at 0? I mean, I can't visualize the two parts but I think I get the concept.

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