Question

If the velocity v of a particle varies as the square of its displacement x then the acceleration varies as

• A. $x^2$
• B. $x^3$
• C. $v^2$
• D. $v^3$

Answer 1

The correct option is B $x^3$

acceleration=$a=$$\frac{vdv}{dx}$

$a=$$k{x}^2\times x$

$a$ is proportional to $x^3$