Question

The displacement 'X' of a particle moving along a straight line at time t is given by
$ x={{a}_{0}}+{{a}_{1}}t+{{a}_{2}}{{t}^{2}} $
The acceleration of the particle is:

• A. $ 4{{a}_{2}} $
• B. $ 2{{a}_{2}} $
• C. $ 2{{a}_{1}} $
• D. $ {{a}_{2}} $

Answer 1

Answer: (B)

Acceleration is known as rate of change of velocity.
The given equation is
$ x={{a}_{0}}+{{a}_{1}}t+{{a}_{2}}{{t}^{2}} $
First differentiate the above equation with respect to t, to obtain velocity that is, $ v=\frac{dx}{dt} $
(velocity is rate of change of displacement)
$ \therefore $ $ v=\frac{dx}{dt}={{a}_{1}}+2{{a}_{2}}t $
$ \left( \text{using}\,\frac{d}{dx}{{x}^{n}}=n\,{{x}^{n-1}} \right) $
Then second differentiation of above equation gives acceleration i.e.,
$ a=\frac{dy}{dt}=2{{a}_{2}} $