Text A particle moves along a line with position law s(t)=t3−6t2+9ts(t)=t^{3}-6t^{2}+9ts(t)=t3−6t2+9t (space in metres, time in seconds). Find the velocity and acceleration at t=2 st=2\text{ s}t=2 s and the instants at which the velocity vanishes. Solution Velocity: v(t)=s′(t)=3t2−12t+9=3(t−1)(t−3)v(t)=s'(t)=3t^{2}-12t+9=3(t-1)(t-3)v(t)=s′(t)=3t2−12t+9=3(t−1)(t−3). Acceleration: a(t)=v′(t)=6t−12a(t)=v'(t)=6t-12a(t)=v′(t)=6t−12. At t=2t=2t=2: v(2)=12−24+9=−3 m/sv(2)=12-24+9=-3\text{ m/s}v(2)=12−24+9=−3 m/s and a(2)=12−12=0 m/s2a(2)=12-12=0\text{ m/s}^2a(2)=12−12=0 m/s2. The velocity vanishes at t=1 st=1\text{ s}t=1 s and t=3 st=3\text{ s}t=3 s. v(2)=−3 m/s;a(2)=0 m/s2;v=0 at t=1,3 s\boxed{v(2)=-3\ \text{m/s};\quad a(2)=0\ \text{m/s}^2;\quad v=0\ \text{at }t=1,3\ \text{s}}v(2)=−3 m/s;a(2)=0 m/s2;v=0 at t=1,3 s