1.

A point moves along a circle with a velocity `v=t//2`. Find the acceleration of the point at the moment when it has covered a quarter circle from the beginning of motion.

Answer» `v=(t)/(2)`
`a_(t)=(dv)/(dt)=(1)/(2)` (constant)
Time taken by the particle to complete a half-circle
`s=ut+(1)/(2)a_(t)t^(2)`
`(piR)/(2)=0+(1)/(2)xx(1)/(2)t^(2)impliest^(2)=2piR`
`a_(c)=(v^(2))/(R )=((t//2)^(2))/(R )=(t^(2))/(4R)=(2piR)/(4R)=(pi)/(2)`
the resultant acceleration when the particle has covered a quarter circle
`a=sqrt(a_(c)^(2)+a_(t)^(2))=sqrt(((pi)/(2))^(2)+((1)/(2))^(2))=(1)/(2)sqrt(pi^(2)+1)`


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