A particle moves in a circle with speed v varying with time as v(t) = 2t. The total acceleration of the particle after it completes 2 rounds of cycle is

Options

A. `16pi`
B. `2 sqrt(1 + 64 pi^2)`
C. `2 sqrt(1 + 49 pi^2)`
D. `14pi `

Answer

B

Explanation

Intantaneous velocity of particle in circular motion is
`v = r omega = at ` v (let v = at)
Then , `omega = (d theta)/(dt) = (at)/(r )`
so , ` int_(0)^(2pi n ) theta = int_(0)^(t) (at)/(r ) dt `,
n = number of rounds = 2 (given)
`rArr 2pi n = (at^2)/(2r) rArr t^2 (4pi nr)/(a)`
Radial acceleration , `a_r = (v^2)/(r) = (a^2 t^2)/(r)`
`rArr a_r = (a^2)/(r) xx (4pi n r)/(a)`
` rArr a_r = 4pi n a `
Tangential acceleration , `a_t = (dv)/(dt) = a`
`therefore ` Toal acceleration ,
` a = sqrt(a_t^2 + a_r^2) = sqrt(a^2 + (4pi n a )^2)`
` = a sqrt(1 + (4pi n)^2)`
` = 2sqrt(1 + 64pi^2)` [given a = 2, n = 2]