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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)#