6. Suppose that the distance of a particle in motion at any time t is denoted and defined by S=2t3-4t2+2t where S is in metres.

Question

1. Find the velocity of this particle at any time t.                                         
2. Give the acceleration of the particle when t=2                                               
3. Find the time(s) when the particle is at rest.                                                    

What distance has the particle covered when it first comes to rest?  

Answer 1

(i) Velocity is the rate of change of distance ds/dt:

ds/dt=6t²-8t+2 m/s

(ii) Acceleration is rate of change of velocity, d²s/dt²=12t-8, which is 24-8=16m/s² when t=2.

(iii) Particle is at rest when velocity=0, so 6t²-8t+2=0=3t²-4t+1=(3t-1)(t-1), so t=⅓ and 1 second when the particle is at rest. Distance for these times is s(⅓)=2/27-4/9+2/3=(2-12+18)/27=8/27 m; and s(1)=2-4+2=0. So the particle is at rest at distances of 8/27 m and 0 m.