Given that v=4t3−2t x=∫vdt,x=t4−t2+C, at t=0,x=0 ⇒ C=0 When particle is 2m away from the origin, then 2=t4−t2⇒t4−t2−2=0 ⇒ (t2−2)(t2+1)=0 ⇒ t=√2s a=
dv
dt
=
d
dt
(4t3−2t) a=12t2−2 For t=√2‌sec a=12×(√2)2−2=22m∕s2