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
$v_{1}=\frac{u+u+a t_{1}}{2}=u+\frac{a}{2} t_{1} \dots$(i)
$v_{2} =\frac{u+a t_{1}+u+a\left(t_{1}+t_{2}\right)}{2}=u+a t_{1}+\frac{a}{2} t_{2} \dots$(ii)
$v_{3} =\frac{u+a\left(t_{1}+t_{2}\right)+u+a\left(t_{1}+t_{2}+t_{3}\right)}{2} $
$=u+a t_{1}+a t_{2}+\frac{a}{2} t_{3} \dots$(iii)
Subtract (i) from (ii), we get;
$v_{2}-v_{1}=\frac{a}{2}\left(t_{1}+t_{2}\right) \dots$(iv)
Subtract (ii) from (iii), we get;
$v_{3}-v_{2}=\frac{a}{2}\left(t_{2}+t_{3}\right) \dots$(v)
Divide (iv) by (v), 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)}$