Question

Velocity of a moving particle varies with time given by v=t^2 +2t +2 . If position of the particle at t=0 is 2m then position of the particle at t=2 sec will be
a) 32/3m
b) 10m
c)38/3m
d)40/3m

Answer 1

In the attachment I have answered this problem         Concept:   Rate of change of displacement is velocity.          That is   Velocity v = dx/dt             See the attachment for detailed solution.  

Answer 2

Velocity of the particle = t² + 2t + 2

dx/dt = t² + 2t + 2

⇒  dx =  (t² + 2t + 2)dt.

Integrating both the sides of the equation.

x = t³/3  + t² + 2t + c

where c is the constant of the Integration.

Solving for c,

at time t = 0 the displacement is 0,

∴ 2 = 0 + 0 + 0 +  c

∴   ║ c = 2 ║

∴ Position of the Particle is given by,

x =  t³/3  + t² + 2t + 2

⇒ x = (2)³/3 + 4 + 2 + 4

⇒ x = 8/3 + 10

∴ x = (8 + 30)/3

∴ x = 38/3 m.

Hence, answer is option (c).

Hope it helps.