Question

The displacement of a particle is proportional to the cube of time elapsed. How does the acceleration of the particle depends on time obtained

• A. $a \propto t^{2}$
• B. $a \propto 2t$
• C. $a \propto t^{3}$
• D. $a \propto t$

Answer 1

Answer: (D)

$a \propto t$

Answer 2

$x \propto t^{3}$ $\therefore x = Kt^{3}$

$\Rightarrow v = \frac{dx}{dt} = 3\mspace{6mu} Kt^{2}$and $\Rightarrow a = \frac{dv}{dt} = 6\mspace{6mu} Kt$ i.e. $a \propto t$