question_answer

The displacement x of a particle along a straight line at time t is given by\[~\text{x }=\alpha -\beta \text{t }+\gamma {{\text{t}}^{\text{2}}}\]. Find the acceleration of the particle.

Answer:

                Let x be the displacement at time t of an object in motion. Given, \[\text{x}=\text{k}{{\text{t}}^{\text{3}}}\], where k is a consant of proportionality. Velocity of object, \[\text{V}=\frac{dx}{dt}=\text{3k }{{\text{t}}^{\text{2}}}\]and acceleration of object,\[\text{A}=\frac{dV}{dt}=\text{3k x 2t }=\text{ 6kt}\]. i.e.\[\text{A}\propto \text{t}\]. It means acceleration \[\propto \] time.