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A particle moves in a circle of rad...

A particle moves in a circle of radius 2 m at a second
what is its resultant (total ) acceleration at time ` t= 1 ` s ?

A

`8 m//s^(2)`

B

` 2 sqrt(2)m//s^(2) `

C

` 5 sqrt(5)m//s^(2)`

D

`4sqrt(5)m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`a_(t ) = (dv ) /(dt ) = (d ) /(dt ) (4t ) = 4 m//s^(2) , ` this remains constant `and a_(c ) = ((v^(2))/(r ))=((4t)^2)/(2) = (16 t^(2) )/(2) = 8 t^(2)`
` therefore ` when `t= 1 s, a_(c ) = 8 m/s ^(2)`
` therefore ` Total acceleration `(a_(R ))` at t=1 s
` =sqrt(a_(t) ^(2)+a_(c )^(2))=sqrt(4^(2)+8^(2))=sqrt(80) = 4sqrt(5) m//s^(2)`
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