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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

A body A starts from rest with an acceleration a1a_1. After 22 seconds, another body B starts from rest with an acceleration a2a_2. If they travel equal distances in the 5th5^{\text{th}} second, after the start of A, then the ratio a1:a2a_1 : a_2 is equal to:

A

5 : 9

B

5 : 7

C

9 : 5

D

9 : 7

Step-by-Step Solution

  1. Formula for Distance in n-th Second: The distance traveled by a body in the nn-th second of its motion is given by Sn=u+a2(2n1)S_n = u + \frac{a}{2}(2n - 1), where uu is the initial velocity and aa is the acceleration .
  2. Motion of Body A: Starts from rest: uA=0u_A = 0. Acceleration: a1a_1. Time interval: 5th5^{\text{th}} second (from t=4t=4 to t=5t=5s). Distance SA=0+a12(2×51)=9a12S_{A} = 0 + \frac{a_1}{2}(2 \times 5 - 1) = \frac{9a_1}{2}.
  3. Motion of Body B: Starts from rest: uB=0u_B = 0. Acceleration: a2a_2.
  • Body B starts 2 seconds after A. Therefore, the 5th5^{\text{th}} second from the start of A corresponds to the interval between t=4t=4 and t=5t=5 seconds on A's clock. Since B started at t=2t=2, this interval corresponds to the time from t=(42)=2t' = (4-2) = 2s to t=(52)=3t' = (5-2) = 3s on B's clock. This is the 3rd3^{\text{rd}} second of B's motion.
  • Distance SB=0+a22(2×31)=5a22S_{B} = 0 + \frac{a_2}{2}(2 \times 3 - 1) = \frac{5a_2}{2}.
  1. Equating Distances: Given SA=SBS_{A} = S_{B}. 9a12=5a22\frac{9a_1}{2} = \frac{5a_2}{2} 9a1=5a29a_1 = 5a_2 a1a2=59\frac{a_1}{a_2} = \frac{5}{9}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOTION IN A STRAIGHT LINE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOTION IN A STRAIGHT LINEstartsaccelerationsecondsanotherstarts

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