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(B) Let the acceleration of the particle be $x$ and the initial velocity be $u=0$.For the first interval of time $p$,the distance covered is $a = \fra

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$\frac{2(bq - ap)}{pq(p+q)}$

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(B) Let the acceleration of the particle be $x$ and the initial velocity be $u=0$.For the first interval of time $p$,the distance covered is $a = \frac{1}{2} x p^2$.For the total time $(p+q)$,the total distance covered is $a+b = \frac{1}{2} x (p+q)^2$.From the first equation,$x = \frac{2a}{p^2}$.Substituting this into the second equation: $a+b = \frac{1}{2} (\frac{2a}{p^2}) (p+q)^2 = a \frac{(p+q)^2}{p^2}$.This approach is complex. Let's use the average velocity method.Velocity at time $p$ is $v_1 = xp$.Distance $a = \frac{1}{2} xp^2$.Distance $b$ is covered in time $q$ starting with velocity $v_1$. So,$b = v_1 q + \frac{1}{2} x q^2 = (xp)q + \frac{1}{2} x q^2 = xpq + \frac{1}{2} x q^2$.We have $a = \frac{1}{2} xp^2 \implies x = \frac{2a}{p^2}$.Substitute $x$ in $b$: $b = (\frac{2a}{p^2})pq + \frac{1}{2} (\frac{2a}{p^2}) q^2 = \frac{2aq}{p} + \frac{aq^2}{p^2} = \frac{2aqp + aq^2}{p^2} = \frac{aq(2p+q)}{p^2}$.Thus,$x = \frac{2(bp - aq)}{pq(p+q)}$ is not the standard form. Let's re-evaluate: $b = \frac{1}{2}x(p+q)^2 - \frac{1}{2}xp^2 = \frac{1}{2}x(q^2+2pq)$.$x = \frac{2b}{q^2+2pq} = \frac{2b}{q(q+2p)}$.

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

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