A particle moving with uniform acceleration along a straight . . .

Question : A particle moving with uniform acceleration along a straight line covers distance a and b in successive intervals of p and q seconds. The acceleration of the particle is
(a) pq(p+q)/2(bp-aq)

(b) 2(aq-bp)/pq(p-q)

(d) 2(bp-aq)/pq(p+q)

Doubt by Suhani

Solution :

It is clearly written that the body is moving with uniform acceleration. It means that we can use equation of motion because they are also applicable for a body moving with uniform acceleration.

The body covers a distance of 'a' in time interval 'p'. It is nowhere written that the body starts from rest hence its initial velocity u should not be equal to zero.

Using 2nd equation of motion

s=ut+½at²
a=up+½ap² — (1)

Now, after p seconds the velocity of body will change. The final velocity of above case will become the initial velocity of the next case.

Using 1st equation of motion

v=u+at
v=u+ap

Now this velocity will become the initial velocity of the next case.

Now, as per the question, the body has covered a distance of 'b' in time interval 'q'

Again using 2nd equation of motion

s=ut+½at²
b=(u+ap)q+½a(q)²

b=uq+apq+½aq² — (1)
After solving equation (1) and (2)

We get

a = 2(bp-aq)/pq(p+q)

Hence, d) 2(bp-aq)/pq(p+q), would be the correct options.