Question

the acceleration of a particle a =3t square plus two t plus 4 where t is in sec acceleration is in mper sec square find the velocity at t = 2 sec

Answer 1

Given,
a = 3t² +2t + 4

concept :- velocity is area bonded regoin of acceleration - time function.

dv/dt = 3t² + 2t + 4
dv = (3t² + 2t + 4)dt
Integrate
Vf - Vo = 3/3 (2³ - 0) + 2/2(2² - 0) + 4(2 - 0)
Vf - Vo = 8 + 4 + 8 = 20
if intial velocity = 0
then,
Vf = 20 m/s

Answer 2

Given, 
a = 3t² +2t + 4 
⇒dv/dt = 3t² + 2t + 4 
⇒dv = (3t² + 2t + 4)dt 
===================================
Vf - Vo = 3/3 (2³ - 0) + 2/2(2² - 0) + 4(2 - 0) 
Vf - Vo = 8 + 4 + 8 = 20 
Let the intial velocity = 0
 ∴Vf = 20 m/s