The speed of a particle changes from `sqrt(5) ms^(-1)` to ` 2sqrt(5) ms^(-1)` in a time `t`. The magnitude of change in its velocity is `5 ms^(-1)` the angle between the initial and final velocities of the particle is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

D

Explanation

Given, `v_(i) = sqrt(5) ms^(-1) , v_(f) = 2sqrt(5) ms^(-1) ` and
`Delta v = 5 ms^(-1)`
Since, both `v_(i) and v_(f)` are extreme speeds, i.e art `t= 0 and t= t`.
So, they can be considered as magnitude of the
velocities at time ,`t = 0 and t = t.`
As we know that
`R^(@) = A^(2) + B^(2) + 2 AB cos theta `
Hence, the angle between the velocities,
`cos theta = (Delta v^(2) - v_(i) ^(2) - v_(f) ^(2) )/(2v_(i) v_(f))`
Putting the given values, we get
`cos theta = ((5)^(2) - (sqrt(5))^(2) - (2sqrt(5))^(2))/(2(sqrt(5))(sqrt(5)))`
`rArr cos theta =(25-5 -20 )/10 = 0/10 = 0 rArr theta = 90^(@)`
Hence, the correct option is (d).