Question

A point moves with uniform acceleration and $ v_1 $ , $ v_2 $ and $ v_3 $ denote the average velocities in the three successive intervals of time $ t_1 $ , $ t_2 $ and $ t_3 $ . Which of the following relation is correct?

• A. $ (v_1 - v_2) : (v_2 - v_3) = (t_1 - t_2) : (t_2 + t_3) $
• B. $ (v_1 - v_2) : (v_2 - v_3) = (t_1 + t_2) : (t_2 + t_3) $
• C. $ (v_1 - v_2) : (v_2 - v_3) = (t_1 - t_2) : (t_2 - t_3) $
• D. $ (v_1 - v_2) : (v_2 - v_3) = (t_1 + t_2) : (t_2 - t_3) $

Answer 1

Answer: (B)

Let $u$ be the initial velocity
$\therefore \, v_{1}=u+a t_{1}, v_{2}=u+a\left(t_{1}+t_{2}\right)$
and $v_{3}=u+a\left(t_{1}+t_{2}+t_{3}\right)$
Now $v_{1}=\frac{u+v_{1}}{2}=\frac{u+\left(u+a t_{1}\right)}{2}=u+\frac{1}{2} a t_{1}$
$v_{2}=\frac{v_{1}+v_{2}}{2}=u+a t_{1}+\frac{1}{2} a t_{2}$
$v_{3}=\frac{v_{2}+v_{3}}{2}=u+a t_{1}+a t_{2}+\frac{1}{2} a t_{3}$
So $\left(v_{1}-v_{2}\right)=-\frac{1}{2} a\left(t_{1}+t_{2}\right)$
and $\left(v_{2}-v_{3}\right)=-\frac{1}{2} a\left(t_{2}+t_{3}\right)$
$\therefore \left(v_{1}-v_{2}\right):\left(v_{2}-v_{3}\right)=\left(t_{1}+t_{2}\right):\left(t_{2}+t_{3}\right)$