A particle is moving in a circle of radious R with constant speed. The time period of the particle is T = 1s. In a time interval `(T)/(6)`, if the difference between average speed and the magnitude of average velocity of the particle is 2m/s, then report `(49R)/(50)`, where R is the radius of the circle (in metres) (take `pi = 3.14` )

Answer

7

Explanation

`(2pi)/(omega) 1s`
`omega = 2 pi` rad/s
`Delta theta = omega Delta t = 2 pi xx (1)/(6) = (pi)/(3)` rad

avg-speed `= 2 R pi`
avg. velocity `= ("displacement")/("time") = (Deltax)/(Delta t)`
`vec(x)_(i) = R hat(j)`
`vec(x)_(f) = R cos ((pi)/(3)) hat(j) + R sin ((pi)/(3)) hat(i)`
`= R [(1)/(2) hat(j) + (sqrt3)/(2) hat(i)]`
`Delta vec(x) = - (R )/(2) hat(j) + (sqrt3)/(2) R hat(i)`

`|Delta vec(x)| = R sqrt((1)/(4) + (3)/(4)) = R`
or `2 theta + (pi)/(3) = pi`
`theta = (pi)/(3)`
equilateral triangle
So `|Delta vec(x)| = R`
`|avg. "velocity"| = (R )/(1//6) = 6R`
`6.28 R - 6R = 0.28 R = 2`
`R = (2)/(28) xx 100 = (50)/(7)`