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A point moves along a circle with a spee...

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)` .

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The correct Answer is:
B
Let velocity of the particle be, `v=(ds)/(st)=kt` or `int_(0)^(s)ds=kint_(0)^(t)tdt` implies :. `s=(1)/(2)kt^(2)`
For completion of `nth` fraction of circle,
`s=2pirn=(1)/(2)kt^(2)` or `t^(2)=(4pinr)//k`
Tangential acceleration `=a_(t)=(dv)/(dt)=k`
Normal acceleration `=a_(n)=(v^(2))/(r)=(k^(2)t^(2))/(r)`
Substituting the value of `t^(2)` from Eq. (i), we have
or `a_(n)=4pirk`
:. `a=sqrt((a_(t)^(2)+a_(n)^(2))=[k^(2)+16pi^(2)n^(2)k^(2)]^(1//2)`
`=k[1+16pi^(2)n^(2)]^(1//2)`
`=0.50[1+16xx(3.14)^(2)xx(0.10)^(2)]^(1//2)`
`=0.8m//s^(2)`
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