Question

A particle in uniformly accelerated motion travels a, b and c distances in xth, yth, and zth second of its motion, respectively. Then a(y - z) + b(z - x) + c (x - y) =

• A. 1
• B. 0
• C. 2
• D. 3

Answer 1

Answer: (B)

0

Answer 2

Using S_n + u + $\frac{a}{2}$ (2n - 1) we get

a = u + $\frac{a'}{2}$ (2x - 1) …………… (i)

[d = uniform acceleration]

b = u + $\frac{a'}{2}$ (2y - 1)…………… (ii)

c = u + $\frac{a'}{2}$ (2z - 1) …………… (iii)

or a = dx + $\left( u - \frac{a'}{2} \right)$…………….. (iv)

From (iv) ay = dxy + $\left( u - \frac{a'}{2} \right)$y

and az = dxy + $\left( u - \frac{a'}{2} \right)$z

Subtracting, a(y - z)

= d(xy - xz) + $\left( u - \frac{a'}{2} \right)$ (y - z)

Similarly, b(z - x)

= a'(yz – yx) + $\left( u - \frac{a'}{2} \right)$ (x - y)

and c(x - y) = d(zx - yz) + $\left( u - \frac{a'}{2} \right)$ (x - y)

Adding above 3 equations

a(y - z) + b(z - x) + c(x - y)

= a'(xy – xz + yz – yx + xz - yz) + $\left( u - \frac{a'}{2} \right)$

(y – z + z – x + x - y) = 0