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Answer 1 · 2017-05-06T09:34:42

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Referring to the figure above
Change in velocity vector $= {\to v}_{B} - {\to v}_{A}$ $= vecv_B-vecv_A$
$=> Delta vecv=(vecv_B) + (-vecv_A)$

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.