Question

A particle is moving along the $$x$$-axis whose instantaneous speed is givn by $$v^2=108-9x^2$$. The acceleration of the particle is

Answer 1

Correct option is A. $$-9x$$ $$ms^{-2}$$
$$v^2=108-9x^2$$ ..........(i)
$$a=\dfrac{dv}{dt}=\dfrac{dv}{dx}.\dfrac{dx}{dt}=\dfrac { d\left( \sqrt { 108-{ 9x }^{ 2 } } \right) }{ dx } -\frac { dx }{ dt } $$
$$a=\dfrac { 1\left( -18x \right) }{ 2\sqrt { 108-{ 9x }^{ 2 } } } .\sqrt { 108-{ 9x }^{ 2 } } =-9x\quad { ms }^{ 2 }$$
Alternative: Differentiating w.r.t x, we get
$$2v\dfrac{dv}{dx}=-18x$$
$$\Rightarrow \dfrac{vdv}{dx}=-9x$$ $$\Rightarrow a=-9x$$ $$\left( \because v\dfrac { dv }{ dx } =a \right) $$