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Answer 1 · 2017-05-19T08:02:48

$v_c=sqrt((v_2^2-v_1^2)/2)$

we have

$ArarrB$

$u=v_1" " v=v_2" let the distance be "s" and accln "a$

using $" "v^2=u^+2as$

$color(blue)(v_2^2=v_1^2+2as)--(1)$

$ArarrC$

$u=v_1" " v=v_c" the distance is "s/2" and accln "a$

using $" "v^2=u^+2as$

$v_c^2=v_1^2+2as/2$

$=>color(blue)(v_c^2=v_1^2+as)--(2)$

$eqn (2) xx2" and rearrange"$

$2v_c^2=2v_1^2+2as$

$color(red)(2v_c^2-2v_1^2=2as)$

substitute into $" "(1)$

$v_2^2=v_1^2+color(red)(2v_c^2-2v_1^2)$

now rearrange for $" "v_c$

$v_2^2=2v_c^2-v_1^2$

$v_c^2=(v_2^2+v_1^2)/2$

$v_c=sqrt((v_2^2+v_1^2)/2)$