Question

A particle moves in a straight line with constant acceleration a. The displacements of particle from origin in times $t_1,t_2$ and $t_3$ are $s_1,s_2$ and $s_3$ respectively. If time are in AP with common difference d and displacements are in GP, then prove that $a=\frac{{\left(\sqrt{s_1-\sqrt{s_3}}\right)}^2}{d^2}$

Answer 1

The particle started from rest at origin i.e $u=0$

Using $s=ut+\frac{a{t}^2}{2}$ where $u=0$

We get $s_1=\frac{a{t}_1^2}{2}$, $s_2=\frac{a{t}_2^2}{2}$ and $s_3=\frac{a{t}_3^2}{2}$

As $s_1,$ $s_2$ & $s_3$ are in GP $⟹$ $s_2=\sqrt{s_1{s}_3}$

Also $t_1,$ $t_2$ & $t_3$ are in AP $⟹2t_2=(t_1+t_3)$ and $2d=t_3-t_1$

Now RHS $\frac{(\sqrt{s_1}-\sqrt{s_3}{)}^2}{d^2}=\frac{s_1+s_3-2\sqrt{s_1{s}_3}}{d^2}=\frac{s_1+s_3-2s_2}{d^2}$

OR RHS $=\frac{a}{2}\times \frac{t_1^2+t_3^2-2t_2^2}{d^2}=\frac{a}{2}\times \frac{t_1^2+t_3^2-2\times \frac{(t_1+t_3{)}^2}{4}}{d^2}$

OR RHS $=\frac{a}{4d^2}\times (t_3-t_1{)}^2=\frac{a}{4d^2}\times (2d{)}^2$ $⟹$ RHS $=a$