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A particle travels along the arc of a ci...

A particle travels along the arc of a circle of radius r . Its speed depends on the distance travelled l as `v=asqrtl` where a is a constant . The angel `alpha` between the vectors of total acceleration and the velocity of the particle is

A

`tan^(-1) ((pi)/(4))`

B

`tan^(-1) ((pi)/(2))`

C

`tan^(-1) (pi)`

D

`tan^(-1)(2pi)`

Text Solution

Verified by Experts

The correct Answer is:
C
Given `v = a_(0)t`
`a_(t) = (dv)/(dt) = a_(0)`
Now, `v^(2) = u^(2) + 2a_(t)S`
`= 0 + 2a_(0) (piR)/(2)` ,
`v^(2) = pia_(0)R`
So, `a_(n) = (v^(2))/(R ) = pia_(0)`
The angle between the velocity vector and the acceleration vector is
`phi = tan^(-1) ((a_(n))/(a_(t))) = tan^(-1)((pia_(0))/(a_(0))) = tan^(-1)(pi)`
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