Question

A particle moving with uniform acceleration has average velocities $v_1$, $v_2$ and $v_3$ over the successive intervals of time $t_1$, $t_2$ and $t_3$ respectively. The value of $\frac{\left(v_1-v_2\right)}{\left(v_2-v_3\right)}$ will be

• A. $\frac{t_1-t_2}{t_2-t_3}$
• B. $\frac{t_1-t_2}{t_2+t_3}$
• C. $\frac{t_1+t_2}{t_2-t_3}$
• D. $\frac{t_1+t_2}{t_2+t_3}$

Answer 1

Answer: (D)

Let $u$ be initial velocity and $a$ be uniform acceleration.

Average velocities in the intervals from $0$ to $t_1$, $t_1$ to $t_2$ and $t_2$ to $t_3$ are
$v_1=\frac{u+u+a{t}_1}{2}=u+\frac{a}{2}{t}_1\dots \left(i\right)$
$v_2=\frac{u+a{t}_1+u+a\left(t_1+t_2\right)}{2}=u+a{t}_1+\frac{a}{2}{t}_2\dots \left(ii\right)$
$v_3=\frac{u+a\left(t_1+t_2\right)+u+a\left(t_1+t_2+t_3\right)}{2}$
$=u+a{t}_1+a{t}_2+\frac{a}{2}{t}_3\dots \left(iii\right)$
Subtract $\left(i\right)from\left(ii\right)$, we get
$v_2-v_1=\frac{a}{2}\left(t_1+t_2\right)\dots \left(iv\right)$
Subtract $\left(ii\right)$ from $\left(iii\right)$, we get
$v_3-v_2=\frac{a}{2}\left(t_2+t_3\right)\dots \left(v\right)$
Divide $\left(iv\right)$ by $\left(v\right)$, we get
$\frac{v_2-v_1}{v_3-v_2}=\frac{\left(t_1+t_2\right)}{\left(t_2+t_3\right)}$ or $\frac{v_1-v_2}{v_2-v_3}=\frac{\left(t_1+t_2\right)}{\left(t_2+t_3\right)}$