The acceleration of a particle in ms^(-1) is given by a=3t^(2)+2t+2 where timer is in second If the particle starts with a velocity v=2ms^(-1) at t=0 then find the velocity at the end of 2s.

The acceleration of a particle in ms^(-1) is given by a=3t^(2)+2t+2 wh

1.	The acceleration of a particle in ms^(-1) is given by a=3t^(2)+2t+2 where timer is in second If the particle starts with a velocity v=2ms^(-1) at t=0 then find the velocity at the end of 2s.
Answer» SOLUTION :`a=(dv)/(dt)=(3T^(2)+2t+2)dt` `dv=(3t+2t+2)dt` `intdv=int(3t^(2)+2t+2)dt` `v=t^(3)+t^(2)+2t+C` `c=2m//s,v=18m//s` at t=2s