A particle `A` is moving in `xy` plane with the constant speed of `2pi m//s` along the path `x^(2)+y^(2)+120y=0`. At time `t=0`. When `A` is at the origin, another particle `B` starts moving from origin with constant acceleration is such a way that at time `t=5s` velocities of both the particles are found to be equal. If sense of motion of `A` is clockwise, calculate the average speed of `B` over first five second.

Options

A. `2pi m//s`
B. `pi m//s`
C. `2 m//s`
D. `1 m//s`

Explanation

`60theta=10pi`
`theta=pi//6`
`vecV_(A)=-v_(0)cos30^(@)hati-v_(0)sin30^(@)hatj`
`=-(v_(0))/(2)[sqrt(3)hati+hatj]`
`rArrvecV_(B)=vecV_(A)=-(V_(0))/(2)[sqrt(3)hati+hatj]`
`veca_(B)=(-V_(0))/(2)[sqrt(3)hati+hatj]`
Since particle starts moving with uniform acceleration so it will move on a straight line
`vecS_(B)=vecU_(B)t+(1)/(2)veca_(B)t^(2)=0+(V_(0))/(20)[sqrt(3)hati+hatj]xx25`
`|vecS_(B)|=` Distance travelled by particle `B=(2pi)/(20)xx25xx2=5pi`
Average speed of particle `B` over first five second `(5pi)/(5)=pim//s`
Second method
`:' veca_(B)=`constant, to
`0 lt V_(B) lt 5sec = |(vecV_(B(a))+vecV_(B(5)))/(2)|=(V_(0))/(2)=pim//s`