Question

A point moves along a circle with a speed v=kt , where K=0.5m//s^(2) Find the total acceleration of the point the momenet when it has covered the n^(th) Fraction of the circle after the beginging of motion, where n=(1)/(10) .

Answer 1

Hence the total acceleration of the point is a =0.8 m/s^2

Explanation:

Let velocity of the particle be

v = ds / st = kt

or ∫s - 0 ds = k∫t - 0 tdt implies :. s = 1 / 2kt^2

For completion of nth fraction of circle,

s = 2πrn  = 1 / 2kt^2

or t^2 = (4πnr) / k

Tangential acceleration = at = dv / dt = k

Normal acceleration = an = v^2 r = k^2t^2 / r  

Substituting the value of t^2 from Eq. (i), we have

or an = 4πrk

a = (a^2 t + a^2n)

= √[k^2 + 16π^2 n^2 k^2]^1/2

a = k[1 + 16π^2 n^2]^1/2  

a = 0.50[1 +16 × (3.14)^2 × (0.10)^2]^1/2

=0.8 m/s^2

Hence the total acceleration of the point is a =0.8 m/s^2