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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

If a freely falling body travels in the last second a distance equal to the distance travelled by it in the first three seconds, the time of the travel is:

A

6 sec

B

5 sec

C

4 sec

D

3 sec

Step-by-Step Solution

  1. Analyze the Motion: The body starts from rest (u=0u=0) and undergoes free fall with acceleration gg. Let the total time of flight be tt seconds. The 'last second' corresponds to the tt-th second.
  2. Distance in First 3 Seconds: Using the equation s=ut+12at2s = ut + \frac{1}{2}at^2: S3=0+12g(3)2=9g2S_3 = 0 + \frac{1}{2}g(3)^2 = \frac{9g}{2} .
  3. Distance in Last (tt-th) Second: The distance covered in the nn-th second is given by Dn=u+a2(2n1)D_n = u + \frac{a}{2}(2n - 1). Here n=tn=t: Dt=0+g2(2t1)D_t = 0 + \frac{g}{2}(2t - 1)
  4. Equate and Solve: The problem states Dt=S3D_t = S_3. g2(2t1)=9g2\frac{g}{2}(2t - 1) = \frac{9g}{2} 2t1=92t - 1 = 9 2t=10    t=5 s2t = 10 \implies t = 5 \text{ s}

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 LINEfreelyfallingtravelsseconddistance

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