The correct option is B 12 m/s
We know that the average speed is given by ¯v=Total displacementTotal time taken
Also, total displacement is given by
xf−xi=∫t0(18−6t)dt=[18t−3t2]t0=18t−3t2 m
Let particle be at origin at t=0⇒ xi=0
and, at t=0,xf=0
also, at t=10,xf=18×10−3×102=−120 m
and total time taken is 10 sec
Thus, ¯v=−12010=−12 m/s
Hence, the magnitude of the average velocity is 12 m/s