In a circular motion of radius 3 cm ...

In a circular motion of radius 3 cm the distance (s) travelled by a body along the circumference bears the relationship `s= ct ^(3) ` with time (t) where c= 0.1 `m//s^(3) ` . What are the tangential and centripetal acceleration when its linear speed is 0.3 m/s?

`0.2 m//s^(2) , 3 m//s ^(2)`

`2 m//s ^(2) , 0.6 m//s^(2)`

`0.6 m//s^(2) , 3 m//s ^(2)`

`2 m//s^(2) ,4m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:

`s= ct^(3)`
` therefore v= (ds)/(dt) = 3 ct^(2)`
` and a = ( dV) /( dt) = 6 ct `
when `v=0.3 m//s , ` then from (1)
`0.3 =0.1 xxt^(2) " " therefore t=1 s`
` therefore ` The tangential accleration `a_(T ) = 6ct = 6 xx0.1 xx1`
`= 0.6 m//s ^(2)`
and C.P acceleration `, a_(c ) = (v^2)/( r ) = (0.3xx0.3)/( 0.03)= 3 m//s^(2)`

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