Question

If a, b, c, be the distance travelled by a particle moving with a constant acceleration during the lth, mth and nth seconds of its motion respectively. Show that a(m - n) +b(n - l) +c (l + m) = 0.

Answer 1

Let 'u' be the initial velocity of the particle and A be its uniform acceleration. Using this relation.

Subtracting (iii) from (ii) we have

Subtracting (i) from (iii) we have,

c -a = A/2 (2n - 2l) = A (n -l)

or, b(n - l) = b(c - a)/A = (bc - ab)/A ....(v)

Subtracting (ii) from (i) we have,

( a - b) = A/2 (2l - 2m) = A(l - m)

or, c(l - m ) = c(a - b)/A = (ac -bc)/A ....(vi)

Adding (iv), (v) and (vi) we have,

a(m- n) + b(n - l) + c(l -m)