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?

Question

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?

anonymous · Answer

attachment

anonymous · Answer

specify what part b of the motion is?

roadjester · Answer

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

anonymous · Answer

27/2 and total distance equals 45/2

roadjester · Answer

Uh, how do you figure? I mean, what formula did you use? (kinematics)

anonymous · Answer

$$v ^{2}-u ^{2} = 2aS$$
v=12,u=-6 and a=4
u get total displacement S..

anonymous · Answer

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

roadjester · Answer

I'm guessing u=initial velocity and v=final velocity. I get v, but why u?

roadjester · Answer

And why stop at 0? I mean, I can't visualize the two parts but I think I get the concept.