Displacement Is Not that easy....

For the correct answer we have to check the motion in each second:

Lets have a look, The important thing here is that S here defines the position of the particle at instant t.

when t=0 S=6. So we see initially the particle is not at zero but at 6m.

when t=1 S=16,so distance is 10m.

when t=2 S=22,so distance is 16m.

when t=3 S=24,so distance is 18m.

when t=4 S=22,here the particle has moved back 2 meter,so the distance is 20m.

when t=5 S=16,here the particle has moved back 6 meter again,so the distance is 26m.

So the final answer after 5 seconds (the distance moved) is 26 meter.

The quadratic equation of displacement has a maxima which occurs at $t=3$ .
Displacement at $t=0$ = 6
Displacement at $t=3$ = 24
Displacement at $t=5$ = 16
Total Distance = ( Positive displacement from $t = 0 to 3$ ) + ( Negative displacement from $t = 3 to 5$ ) = 18 + 8 = 26

Take the derivative to get velocity. Integrate the absolute value of the velocity function from 0 to 5.

S=6+12t -(2t t) v=12-4t , v=0 after 3 seconds. a=-4 total distance traversed= 12 3- 1/2 4 (3 3) + 1/2 4* (2*2)=26

s = 6 + 12t - 2t^2, v = 12 - 4t, a = -4. At t = 0 vo = 12. At t = 3 v = 0. So at 3 s, the distance traveled is S3 = 36 - 18 = 18 m. Displacement in 5 s is S5 = 12 x5 + 0.5x(-4)25 = 10. So the total distance traveled is 18 + (18 - 10) = 26 m.

t=0 , s=6 t=1 , s=16 t=2 , s=22 t=3 , s=24 t=4 , s= 22 t=5 , s=16 hence dist moved is 6+ ( 22 - 16) + (24 - 22) + ( 24 - 22) + (24 - 16) = 26 m