A body travels 102.5 m in $\mathrm{n}^{\text {th }}$ second and 115.0 m in $(\mathrm{n}+2)^{\text {th }}$ second. The acceleration is :

(B) $6.25 \mathrm{~m} / \mathrm{s}^{2}$

(A) $9 \mathrm{~m} / \mathrm{s}^{2}$

(C) $12.5 \mathrm{~m} / \mathrm{s}^{2}$

(D) $5 \mathrm{~m} / \mathrm{s}^{2}$

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Kinematics

A body travels 102.5 m in

n^{th}

second and 115.0 m in

(n + 2)^{th}

second. The acceleration is :

(A) $9 m / s^{2}$
(B) $6.25 m / s^{2}$ ✓ Correct
(C) $12.5 m / s^{2}$
(D) $5 m / s^{2}$

Answer: (B)

Solution

Distance in the

n^{th}

second:

s_{n} = u + \frac{a}{2} (2 n - 1)

102.5 = u + \frac{a}{2} (2 n - 1), 115.0 = u + \frac{a}{2} (2 n + 3) .

Subtracting:

12.5 = \frac{a}{2} [(2 n + 3) - (2 n - 1)] = \frac{a}{2} (4) = 2 a \Rightarrow a = 6.25 m/s^{2} .

Option B.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Single correct option

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