Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A body A starts from rest with an acceleration $a_1$ . After $2$ seconds, another body B starts from rest with an acceleration $a_2$ . If they travel equal distances in the $5^{\text{th}}$ second, after the start of A, then the ratio $a_1 : a_2$ is equal to:

5 : 9

5 : 7

9 : 5

9 : 7

Step-by-Step Solution

Formula for Distance in n-th Second: The distance traveled by a body in the $n$ -th second of its motion is given by $S_n = u + \frac{a}{2}(2n - 1)$ , where $u$ is the initial velocity and $a$ is the acceleration .
Motion of Body A: Starts from rest: $u_A = 0$ . Acceleration: $a_1$ . Time interval: $5^{\text{th}}$ second (from $t=4$ to $t=5$ s). Distance $S_{A} = 0 + \frac{a_1}{2}(2 \times 5 - 1) = \frac{9a_1}{2}$ .
Motion of Body B: Starts from rest: $u_B = 0$ . Acceleration: $a_2$ .

Body B starts 2 seconds after A. Therefore, the $5^{\text{th}}$ second from the start of A corresponds to the interval between $t=4$ and $t=5$ seconds on A's clock. Since B started at $t=2$ , this interval corresponds to the time from $t' = (4-2) = 2$ s to $t' = (5-2) = 3$ s on B's clock. This is the $3^{\text{rd}}$ second of B's motion.
Distance $S_{B} = 0 + \frac{a_2}{2}(2 \times 3 - 1) = \frac{5a_2}{2}$ .

Equating Distances: Given $S_{A} = S_{B}$ . $\frac{9a_1}{2} = \frac{5a_2}{2}$ $9a_1 = 5a_2$ $\frac{a_1}{a_2} = \frac{5}{9}$ .

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