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NEET PHYSICSEasy

The initial velocity of a particle is 10 m/sec10 \text{ m/sec} and its retardation is 2 m/sec22 \text{ m/sec}^2. The distance moved by the particle in the 5th second of its motion is:

A

1 m

B

19 m

C

50 m

D

75 m

Step-by-Step Solution

  1. Analyze Motion: The particle has an initial velocity u=10 m/su = 10 \text{ m/s} and acceleration a=−2 m/s2a = -2 \text{ m/s}^2 (retardation). First, check when the particle comes to rest using v=u+atv = u + at. 0=10−2t  ⟹  t=5 s0 = 10 - 2t \implies t = 5 \text{ s}. Since the particle stops exactly at the end of the 5th second, it does not change direction during the 5th second (interval t=4t=4 to t=5t=5). Thus, distance equals displacement.
  2. Formula for nn-th Second: The displacement in the nn-th second is given by Sn=u+a2(2n−1)S_n = u + \frac{a}{2}(2n - 1).
  3. Calculation: Substitute u=10u=10, a=−2a=-2, and n=5n=5: S5=10+−22(2×5−1)S_5 = 10 + \frac{-2}{2}(2 \times 5 - 1) S5=10−1(9)S_5 = 10 - 1(9) S5=10−9=1 mS_5 = 10 - 9 = 1 \text{ m}.
  4. Alternative Method: Calculate position at t=4t=4 and t=5t=5 using x=ut+12at2x = ut + \frac{1}{2}at^2 . x5=10(5)−12(2)(25)=50−25=25 mx_5 = 10(5) - \frac{1}{2}(2)(25) = 50 - 25 = 25 \text{ m}. x4=10(4)−12(2)(16)=40−16=24 mx_4 = 10(4) - \frac{1}{2}(2)(16) = 40 - 16 = 24 \text{ m}. Distance in 5th second =x5−x4=25−24=1 m= x_5 - x_4 = 25 - 24 = 1 \text{ m}.
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