Question

the position of a vector moving along a straight line is given by x=2-5t+t^3 the acceleration of a particle at t=2 s​

Answer 1

Answer:

Explanation:

GIVEN :

X = 2-5T+T³

DIFFERENTIATING:

DX/DT = D/DT(2-5T+T³)

V = -5+3T².

AGAIN DIFFERENTIATING:

A = 6T

ACCELERATION AT T = 2 SEC

A = 6*2

   = 12 M/S².

THEREFORE THE ACCELERATION OF THE PARTICLE A TIME T = 2 SEC IS 12 M/S² .

HOPE THIS HELPS.

Answer 2

Answer:

12

Explanation:

x=2-5t+t^3

after differentiating =-5+3t^2

again diff for acc =6t

6×2=12