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

Question

A particle moving with a uniform acceleration along a straight line covers distance a and b in successive intervals of p and q second. The acceleration of the particle is (A) (p q(p+q)/2(b p-a q)) (B) (2(a q-b p)/p q(p-q)) (C) (b p-a q/p q(p-q)) (D) (2(b p-a q)/p q(p-q))

Tardigrade Admin · Accepted Answer

According to problem, when s=a, t=p ∵ s=u t+(1/2) f t2 (here,f= acceleration) ∴ a=u p+(f p2/2).....(i) For s=b, t=q b=u q+(f q2/2)........(ii) After solving Eqs. (i) and (ii), we get f=(2(a q-b p)/p q(p-q))

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

$\frac{pq ( p + q )}{2 ( b p - a q )}$

$\frac{2 ( a q - b p )}{pq ( p - q )}$

$\frac{b p - a q}{pq ( p - q )}$

$\frac{2 ( b p - a q )}{pq ( p - q )}$

According to problem, when
$s = a, t = p$
$∵ s = u t + \frac{1}{2} f t^{2}$ (here,f= acceleration)
$∴ a = u p + \frac{f p ^{2}}{2}$ .....(i)
For $s = b, t = q$
$b = u q + \frac{f q ^{2}}{2}$ ........(ii)
After solving Eqs. (i) and (ii), we get
$f = \frac{2 ( a q - b p )}{pq ( p - q )}$

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

2(bp−aq)pq(p+q)​

pq(p−q)2(aq−bp)​

pq(p−q)bp−aq​

pq(p−q)2(bp−aq)​