Question

28. If a, b and c be the distances travelled by the body during xth yth and zth second from start, then which of
the following relations is true?
(1) aly - Z) + b(z - x) + c(x - y) = 0
(2) alz - x) + b(x - y) + cly - z) = 0
(3) alx - y) + bly - z) + c(z – X) = 0
(4ax + by + cz = 0​

Answer 1

answer : c(x - y) + a(y - z) + b(z - x) = 0

explanation : we know, distance travelled by body during nth second is ...

$\boxed{\bf{S_n=u+\frac{1}{2}A(2n-1)}}$

where u is initial velocity and A is acceleration of the body.

given, distance travelled during xth second = a

i.e., $S_x=a$

or, u + 1/2 A(2x - 1) = a .....(1)

similarly,

distance travelled during yth second = b

i.e., $S_y=b$

or, u + 1/2 A(2y - 1) = b......(2)

and distance travelled during zth second = c

i.e., $S_z=c$

or, u + 1/2 A(2z - 1) = c......(3)

from equations (1) and (2),

1/2 A(2x - 2y) = (a - b)

or, A(x - y) = (a - b)

or, Ac(x - y) = ac - bc ......(4)

from equations (2) and (3),

A(y - z) = (b - c) .

Aa(y - z) = ab - ca ......(5)

from equations (3) and (1),

A(z - x) = (c - a)

Ab(z - x) = bc - ab ........(6)

now adding equations (4), (5) and (6),

Ac(x - y) + Aa(y - z) + Ab(z - x) = ac - bc + ab - ca + bc - ca = 0

or, c(x - y) + a(y - z) + b(z - x) = 0

Answer 2

Answer:

a(y−z)+b(z−x)+c(x−y)=0.

Explanation: