If a, b, c be the space described in the pth, qth and rth seconds by a particle with a given velocity and moving with uniform acceleration in a straight line then what is the value of a(q – r) + b(r – p) + c(p – q)?

(a) 0

(b) 1

(c) -1

(d) Can’t be determined

The question was asked during an online exam.

My question is based upon Calculus Application in portion Application of Calculus of Mathematics – Class 12

NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options

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Correct option is (a) 0

For explanation: Let, f be the uniform acceleration and u be the given initial velocity of the moving particle.

From the conditions of problem we have the following equation of motion of the particle:

u + ½(f)(2p – 1) = a ……….(1)

u + ½(f)(2q – 1) = b ……….(2)

u + ½(f)(2r – 1) = c ……….(1)

Thus, a(q – r) + b(r – p) + c(p – q) = [u + ½(f)(2p – 1)](q – r) + [u + ½(f)(2q – 1)](r – p) + [u + ½(f)(2r – 1)](p – q)

= u (q – r + r – p + p – q) + f/2[(2p – 1)(q – r) + (2q – 1)(r – p) + (2r – 1)(p – q)]

= u*0 + f/2[2(pq – rp + qr – pq + rp – qr) – q + r – r + p – p + q]

= f/2*0

= 0