Question

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

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

Answer 1

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

According to the given problem, when
$s=a,t=p$
$\therefore s=ut+\frac{1}{2}c{t}^2$ (where $c=$ acceleration)
or $a=up+\frac{c{p}^2}{2}$ ...$\left(1\right)$
For $s=b,t=q$
$b=uq+\frac{c{q}^2}{2}$ ...$\left(2\right)$
Now by multiplying $(1\times q)$ and $(2\times p)$ and subtracting both the equations we get,

$c=\frac{2(aq-bp)}{pq(p-q)}$
Hence, the correct answer is option (b).