Question

Question #3c164

Answer 1

Referring to the figure above
Change in velocity vector`= vecv_B-vecv_A`
`=> Delta vecv=(vecv_B) + (-vecv_A)`

Angle between two velocity vectors is `=40^@`
Since the person moves with with constant speed `v`, we have
`|vecv_A|=|vecv_B|=v`

Magnitude of Resultant vector `|Delta vecv|= sqrt(v^² + v^² - 2v^²cos40^@)`
`= sqrt(2v^²(1 - cos40^@))`

Using the identity
`cos2x = 1 – 2sin^²(x)`
`=> 1 - cos2x = 2sin^²x`, we get

`|Delta vecv|= sqrt(2v^²(2sin^²20^@)) = 2vsin20^@`

Proved.