Question

A particle moves in a circular path of radius R such that its speed V varies with distance x as we equals to Alpha under root x Where Alpha is a positive constant acceleration of the particle after travelling a distance under root 2 hour is​

Answer 1

it is given that, a particle in a circular path of radius R such that its speed varies as $v=\alpha\sqrt{x}$

so, centripetal acceleration, $a_c$ = v²/R = α²x/R

here, x = s

so, $a_c$ = α²s/R

now, tangential acceleration, $a_t$ = dv/dt = α d(√x)/dt = α/2√x × dx/dt

= α/2√x × v

= α/2√x × α√x

= α²/2

we know, net acceleration in circular path, $a_{net}=\sqrt{a_t^2+a_c^2}$

= $\sqrt{\left(\frac{\alpha^2s}{R}\right)^2+\left(\frac{\alpha^2}{2}\right)^2}$

= $\alpha^2\sqrt{\frac{s^2}{R^2}+\frac{1}{4}}$

hence, acceleration of the particle after traveling a distance s is $\alpha^2\sqrt{\frac{s^2}{R^2}+\frac{1}{4}}$

Answer 2

$\huge\bold\purple{Answer:-}$

so, centripetal acceleration, a_c = v²/R = α²x/R

here, x = s

so, a_c = α²s/R

now, tangential acceleration, a_t = dv/dt = α d(√x)/dt = α/2√x × dx/dt

= α/2√x × v

= α/2√x × α√x

= α²/2

we know, net acceleration in circular path, a_{net}=\sqrt{a_t^2+a_c^2}

= \sqrt{\left(\frac{\alpha^2s}{R}\right)^2+\left(\frac{\alpha^2}{2}\right)^2}

= \alpha^2\sqrt{\frac{s^2}{R^2}+\frac{1}{4}}

hence, acceleration of the particle after traveling a distance s is \alpha^2\sqrt{\frac{s^2}{R^2}+\frac{1}{4}}