A girl riding a bicycle with a speed of ` 5 m//s` to wards Noth direction, observes rain falling vertically down. If she increases her speed to `10 m//s`, rain appeard to meet her at ` 45^@` to the vertical . What is the speed ot the rain ? In what direction does rain fall as observed by a ground based observer ?

Options

A. `5sqrt(2)ms^(-1)`
B. `5 m s^(-1)`
C. `10sqrt(2) m s^(-1)`
D. `10 m s^(-1)`

Answer

A

Explanation

Assume north to be `hati` direction and vertically upwards to be `hatj` direction.
let the velocity of rain be `vecv_(r) = a hati + b hatj`
In the first case, `vecv_(g) ` = velocity of girl ` = 5hati`
`therefore vecv_(rg) = vecv_(r) - vecv_(g) = (a hati + b hatj) - 5 hati = (a-5)hati + bhatj`
Since rain appears to fall vertically downwards,
`therefore a - 5 = 0 or a = 5`
In the second case, `vecv_(g) = 10 hati`
`therefore vecv_(rg) = vecv_(r) -10 hati = (ahati + bhatj) - 10hati = (a -10) hati + bhatj = - 5hati + b hatj`
Since rain appears to fall at `45^(@)` with the vertical,
Since rain appears to fall at `45^(@)` with vertical,
`therefore a = b = -5`
`therefore` The velocity of the rain is `vecv_(r) = 5hati - 5hatj`
Speed of the rain is `|vecv_(r)| = sqrt((5)^(2) + (-5)^(2)) = 5sqrt(2) ms^(-1)`