Question

The velocity time relation of a particle is given by v=(3t2-2t-1)m/s calculate the position and acceleration of the particle when the velocity of the particle is zero. given the initial position of the object is 5m.

Answer 1

a=dv/dt

a=3t^2+2t+2

dv=a*dt

dv=(3t^2+2t+2)dt

integrating both sides we get

V=3xt^3/3+2t^2/2+2t=t^3+t^2+2t+C(constant)

at t=o

v=2m/s

2=0+c

c=2

V=t^3+t^2+2t+2

at t=2

V=8+4+4+2

=18

Therfore V=18m/s

Answer 2

Explanation:

a=dv/dt

a=3t^2+2t+2

dv=a*dt

dv=(3t^2+2t+2)dt

integrating both sides we get

V=3xt^3/3+2t^2/2+2t=t^3+t^2+2t+C(constant)

at t=o

v=2m/s

2=0+c

c=2

V=t^3+t^2+2t+2

at t=2

V=8+4+4+2

=18

Therfore V=18m/s