Question

A body moving along a straight line with uniform acceleration, starts from point $O$ and through points $A,B,C$ and $D$ at distances $x_1,x_2.x_3$ and $x_4$ from $O$. It covers distances $AB,BC$ and $CD$ in equal intervals of time. Then $(x_4-x_1)$ is equal to

• A. $x_2-x_4$
• B. $x_2-x_3$
• C. $3(x_2-x_3)$
• D. $3(x_3-x_2)$

Answer 1

The correct option is D

$3(x_3-x_2)$

let particle crosses $A$, $B$, $C$ & $D$ at time $t_1$, $t_2$, $t_3$ & $t_4$ respectively. let accelereation of body is $a$ then,
$x_1=\frac{1}{2}a{t}_1^2$
$x_2=\frac{1}{2}a{t}_2^2$
$x_3=\frac{1}{2}a{t}_3^2$ and
$x_4=\frac{1}{2}a{t}_4^2$
according to given information
$t_2-t_1=t_3-t_2=t_4-t_3$
Solving these equations we will get the answer