Question

A particle is moving along positive x- axis. Its position varies asx=t3−3t2+12t+20 Initial acceleration of the particle is.

Answer 1

Answer:

$-6ms^{-2}$

Explanation:

$velocity, v =\frac{dx}{dt} =\frac{d}{dt} (t^{3} -3t^{2} +2t +20)= 3t^{2} -6t+12\\acceleration, a= \frac{dv}{dt} =\frac{d}{dt} (3t^{2} -6t+12)=6t-6$

initially at t = 0,

acceleration, a= 6(0) -6 = $-6ms^{-2}$