Question #fa5a6

Question

888 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-19T08:02:48

$v_{c} = \sqrt{\frac{v_{2}^{2} - v_{1}^{2}}{2}}$ $v_c=sqrt((v_2^2-v_1^2)/2)$

we have

$A \to B$ $ArarrB$

$u = v_{1} v = v_{2} let the distance be s and accln a$ $u=v_1" " v=v_2" let the distance be "s" and accln "a$

using $v^{2} = u^{+} 2 a s$ $" "v^2=u^+2as$

$v_{2}^{2} = v_{1}^{2} + 2 a s - - (1)$ $color(blue)(v_2^2=v_1^2+2as)--(1)$

$A \to C$ $ArarrC$

$u = v_{1} v = v_{c} the distance is \frac{s}{2} and accln a$ $u=v_1" " v=v_c" the distance is "s/2" and accln "a$

using $v^{2} = u^{+} 2 a s$ $" "v^2=u^+2as$

$v_{c}^{2} = v_{1}^{2} + 2 a \frac{s}{2}$ $v_c^2=v_1^2+2as/2$

$\Rightarrow v_{c}^{2} = v_{1}^{2} + a s - - (2)$ $=>color(blue)(v_c^2=v_1^2+as)--(2)$

$e q n (2) \times 2 and rearrange$ $eqn (2) xx2" and rearrange"$

$2 v_{c}^{2} = 2 v_{1}^{2} + 2 a s$ $2v_c^2=2v_1^2+2as$

$2 v_{c}^{2} - 2 v_{1}^{2} = 2 a s$ $color(red)(2v_c^2-2v_1^2=2as)$

substitute into $(1)$ $" "(1)$

$v_{2}^{2} = v_{1}^{2} + 2 v_{c}^{2} - 2 v_{1}^{2}$ $v_2^2=v_1^2+color(red)(2v_c^2-2v_1^2)$

now rearrange for $v_{c}$ $" "v_c$

$v_{2}^{2} = 2 v_{c}^{2} - v_{1}^{2}$ $v_2^2=2v_c^2-v_1^2$

$v_{c}^{2} = \frac{v_{2}^{2} + v_{1}^{2}}{2}$ $v_c^2=(v_2^2+v_1^2)/2$

$v_{c} = \sqrt{\frac{v_{2}^{2} + v_{1}^{2}}{2}}$ $v_c=sqrt((v_2^2+v_1^2)/2)$