Question

A particle moves in a straight line with constant acceleration and its distances from the origin $O$ on the line at times $t_1,t_2$ and $t_3$ are $x_1,x_2$ and $x_3$ respectlvely. lf $t_1,t_2,t_3$ form an arithmetic progression whose common difference is $d$ and $x_1,x_2,x_3$ are in geometric progression then the acceleration is

• A. $\frac{\sqrt{x_1}+\sqrt{x_2}}{d^2}$
• B. $\frac{{(\sqrt{x_1}-\sqrt{x_3})}^2}{2d^2}$
• C. $\frac{x_1-x_2}{\sqrt{d}}$
• D. $\frac{{(\sqrt{x_1}+\sqrt{x_2})}^2}{\sqrt{d}}$

Answer 1

Answer: (B)

$\frac{{(\sqrt{x_1}-\sqrt{x_3})}^2}{2d^2}$

By using 2nd equation of motion

S=ut+$\frac{1}{2}a{t}^2$

$x_1=\frac{1}{2}a{t}_1^2$ $⟶\left(1\right)$

$x_2=\frac{1}{2}a{t}_2^2$ $⟶\left(2\right)$

$x_3=\frac{1}{2}a{t}_3^2$ $⟶\left(3\right)$

so, by using (1) and (3)

we get,

$\sqrt{x_1}-\sqrt{x_3}=\sqrt{\frac{a}{2}}(t_1-t_3)$

here, $t_1$ - $t_3$ = 2d

$\sqrt{x_1}-\sqrt{x_3}=\sqrt{\frac{a}{2}}$ * (2d)

a = $\frac{{(\sqrt{x_1}-\sqrt{x_3})}^2}{2d^2}$