A particle moving in a straight line with uniform acceleration is observed to be a distance a from a fixed point initially. It is at distances b, c, d from the same point after n, 2 n, 3 n second. The acceleration of the particle is

Question

A particle moving in a straight line with uniform acceleration is observed to be a distance a from a fixed point initially. It is at distances b, c, d from the same point after n, 2 n, 3 n second. The acceleration of the particle is (A) (c-2 b+a/n2) (B) (c+b+a/9 n2) (C) (c+2 b+a/4 n2) (D) (c-b+a/n2)

Tardigrade Admin · Accepted Answer

As, b-a=u n+(1/2) A n2 ∴ 2 b-2 a=2 u n+A n2 ......(i) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/2c20ba2bc4f6a5b87f29dc7341382aec-.png" /> Again, C - a =u(2 n)+(1/2) A (2 n)2 .......(ii) Subtracting, Eq. (i) in Eq. (ii), we get c-a-2 b+2 a =A n2 A =(c-2 b+a/n2)

Q. A particle moving in a straight line with uniform acceleration is observed to be a distance $a$ from a fixed point initially. It is at distances $b, c, d$ from the same point after $n, 2 n, 3 n$ second. The acceleration of the particle is

$\frac{c - 2 b + a}{n ^{2}}$

$\frac{c + b + a}{9 n ^{2}}$

$\frac{c + 2 b + a}{4 n ^{2}}$

$\frac{c - b + a}{n ^{2}}$

As, $b - a = u n + \frac{1}{2} A n^{2}$
$∴ 2 b - 2 a = 2 u n + A n^{2}$ ......(i)

Again, $C - a = u (2 n) + \frac{1}{2} A (2 n)^{2}$ .......(ii)
Subtracting, Eq. (i) in Eq. (ii), we get
$c - a - 2 b + 2 a = A n^{2}$
$A = \frac{c - 2 b + a}{n ^{2}}$

Q. A particle moving in a straight line with uniform acceleration is observed to be a distance a from a fixed point initially. It is at distances b,c,d from the same point after n,2n,3n second. The acceleration of the particle is

n2c−2b+a​

9n2c+b+a​

4n2c+2b+a​

n2c−b+a​