Question

A particle with uniform retardation along a straight line distances $a$ and $b$ in successive intervals $p$ and $q$ seconds. The acceleration of the particle is

• A. $\frac{2(aq-bp)}{pq(p+q)}$
• B. $\frac{2(bp-aq)}{q(p+q)}$
• C. $\frac{2(aq+bp)}{pq(p+q)}$
• D. $\frac{2(aq-bp)}{pq(p-q)}$

Answer 1

The correct option is A $\frac{2(aq-bp)}{pq(p+q)}$

Let retardation be $f$ and initial velocity be $u$.
For $AB$
$a=up-\frac{1}{2}f{p}^2...\left(i\right)$
$\frac{a}{p}=u-\frac{fp}{2}$
For $AC,a+b=u(p+q)-\frac{1}{2}f(p+q{)}^2...(ii)$
$\frac{a+b}{p+q}=u-\frac{f(p+q)}{2}$

Subtracting (i) and (ii)

$\frac{a+b}{p+q}-\frac{a}{p}=\frac{fp}{2}-\frac{f(p+q)}{2}$
$\frac{pa+pb-ap-aq}{p(p+q)}=-\frac{fq}{2}$
$f=\frac{2(aq-bp)}{pq(p+q)}$