Question

The acceleration of a particle is given by a= 3t2 + 2t + 2 where t is time in sec and a is acc in m/s2 .If initial velocity is 2 m/s, then find the velocity at the end of 2 sec.( 18 m/s)

Answer 1

Explanation:

given a=3t²+2t+2

we know a=d/dt(v)

so adt=dv

INTEGRATING ON BOTH SIDES

=>v=integral of (3t²+2t+2)

=>v={3×(t^3/3)+2(t^2/2)×2t)

=t^3+t^2+2t

at t=2sec

v=8+4+4=16

ANOTHER PROCESS

a=3t^2+2t+2

at t=2sec

a=12+4+2=18

we know

a=(v-u)/t

18=(v-2)/2

v=38m/s.

check it if is it correct or not