Question

the velocity displacement equation of any particle is v=
$\sqrt{180 - 16x} \sqrt{ \sqrt \sqrt{?} } } \times \frac{?}{?}$
where v and x are in SI unit .so find the acceleration of the particle ​

Answer 1

Answer:

$\boxed{\mathfrak{Acceleration = -8 \ m/s^2}}$

Given:

Velocity (v) w.r.t displacement (x) of any particle:

$\rm v = \sqrt{180 - 16x}$

To Find:

Acceleration (a) of the particle

Explanation:

Rate of change of velocity w.r.t time is equal to acceleration:

$\rm \implies a = \frac{dv}{dt} \\ \\ \rm \implies a = \frac{dv}{dt} \times \frac{dx}{dx} \\ \\ \rm \implies a = \frac{dx}{dt} \times \frac{dv}{dx} \\ \\ \rm \frac{dx}{dt} = v : \\ \rm \implies a =v \frac{dv}{dx} \\ \\ \rm v = \sqrt{180 - 16x} : \\ \rm \implies a = (\sqrt{180 - 16x} ) \times \frac{d( \sqrt{180 - 16x} )}{dx} \\ \\ \rm \implies a = \cancel{\sqrt{180 - 16x}} \times \frac{ - 16}{2 \cancel{ \sqrt{180 - 16x} }} \\ \\ \rm \implies a = - \frac{16}{2} \\ \\ \rm \implies a = - 8 \: m/s^2$

$\therefore$

Acceleration (a) of the particle = -8 m/s²

Answer 2

Answer:

a = vdv/dx

= √(180 -16x) × d(√(180 - 16x))/dx

= -16/2

= -8 ms^-2