Question

3.
A particle moves in a circular path of radius R such that its speed v varies with distance s as
$v = \sqrt{s}$
where

$\alpha$
is
positive constant. The acceleration of the particle after traversing a distance s is

​

Answer 1

Answer:

a^2√1/4+s^2/R^2

Explanation:

v=a√s

an=v^2/R=a^2S/R

Answer 2

Answer:

The acceleration of the particle after traversing a distance of s is

α² *√[1/4 + s²/R²].

Explanation:

A particle is given to be moving in a circular path whose radius = R

The speed, v = α √s …… where α is given as a positive constant

Firstly, let’s differentiate the given equation of speed “v” w.r.t time to “t”, so,

dv/dt = d[α √s]/dt  

⇒ dv/dt = ½ * α * (s)^(-1/2) * (ds/dt)

since the v varies with distance s ∴ substituting ds/dt = α √s, we get

⇒ dv/dt =  ½ * α * (1/√s) * α √s

⇒ dv/dt = α² / 2 …… (i)

Now, here we need to find the acceleration of the particle after traversing a distance of s, so we will calculate the total acceleration which is given as,

Total acceleration (vector a) = Tangential acceleration(vector at) + radial acceleration(vector ar)

Where tangential acceleration(at) = dv/dt  and radial acceleration(ar) = v²/R

∴ The magnitude of the total acceleration of the particle is,  

= √[(at)² + (ar)²]  

= √[( dv/dt  )² + (v²/R)²]  

= √[(α²/2)² + {( α √s )² / R}²]

= √[(α⁴/4) + {α²*s / R}²]

= √[{α⁴/4} + {(α²*s)²/(R)²}]

= √[{α⁴ / 4} + {(α*s)²/(R)²}]

= α² * √[1/4 + s²/R² ]