1 \(\dfrac{u^{2}-v^{2}}{\alpha+\beta}\)
2 \(\dfrac{u^{2}+v^{2}}{\alpha+\beta}\)
3 \(\dfrac{u^{2}-v^{2}}{\alpha-\beta}\)
4 \(\dfrac{u^{2}+v^{2}}{\alpha-\beta}\)
Explanation:
Let the distance be \(p\) when velocity is \(u\) and acceleration \(\alpha\)
Let the distance \(q\) when velocity is \(v\) and acceleration \(\beta\)
If \(\omega\) is the angular frequency, then
\(\alpha = {\omega ^2}p{\rm{ and }}\beta = {\omega ^2}p\)
\(\therefore \alpha + \beta = {\omega ^2}(p + q)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
Also, \(u^{2}=\omega^{2} A^{2}-\omega^{2} p^{2}\)
and \(v^{2}=\omega^{2} A^{2}-\omega^{2} q^{2}\)
\( \Rightarrow {v^2} - {u^2} = {\omega ^2}\left( {{p^2} - {q^2}} \right)\)
\({v^2} - {u^2} = {\omega ^2}(p - q)(p + q)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\)
By eqs.(1) and (2), we get
\(\begin{aligned}v^{2}-u^{2} & =(p-q)(\alpha+\beta) \\\therefore p-q & =\dfrac{v^{2}-u^{2}}{\alpha+\beta} \text { or } q-p=\dfrac{u^{2}-v^{2}}{\alpha+\beta}\end{aligned}\)