Question

x-coordinate of a particle moving along this axis is x = (2+t^2 + 2t^3). Here, x is in meres and t in seconds. Find (a) position of particle from where it started its journey, (b) initial velocity of particle and (c) acceleration of particle at t=2s.

Answer 1

Answer:

(A) initial position is at t = 0

      x = ( 2 + 0^2 + 0^3)

      x = 2 m

(B) differentiate to get the velocity

 on differentiation ,  v = 2t + 6t^2

now put t = o,     v = 0 m/s

(C)  now differentiate velocity to get acceleration

on differentiation ,  a = 2 + 12t

put t = 2,      a = 2 + 12× 2

                    a = 26 m/s^2