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If a body having initial velocity zero is moving with uniform acceleration 8 m/sec28 \text{ m/sec}^2, then the distance travelled by it in the fifth second will be:

A

36 metres

B

40 metres

C

100 metres

D

Zero

Step-by-Step Solution

  1. Identify Given Values: Initial velocity, u=0u = 0. Uniform acceleration, a=8 m/s2a = 8 \text{ m/s}^2.
  • Time interval: The 5th second (n=5n = 5).
  1. Method 1: Using the n-th second formula The distance travelled in the nn-th second is given by Sn=u+a2(2n−1)S_n = u + \frac{a}{2}(2n - 1). Substitute the values: S5=0+82(2×5−1)S_5 = 0 + \frac{8}{2}(2 \times 5 - 1) S5=4(10−1)=4×9=36 mS_5 = 4(10 - 1) = 4 \times 9 = 36 \text{ m}.
  2. Method 2: Difference in Displacement Calculate the position at t=5t=5 and t=4t=4 using x=ut+12at2x = ut + \frac{1}{2}at^2. x5=0(5)+12(8)(5)2=4(25)=100 mx_5 = 0(5) + \frac{1}{2}(8)(5)^2 = 4(25) = 100 \text{ m}. x4=0(4)+12(8)(4)2=4(16)=64 mx_4 = 0(4) + \frac{1}{2}(8)(4)^2 = 4(16) = 64 \text{ m}.
  • Distance in 5th second =x5−x4=100−64=36 m= x_5 - x_4 = 100 - 64 = 36 \text{ m}.
  1. Method 3: Galileo's Law of Odd Numbers For a particle starting from rest with uniform acceleration, distances travelled in successive equal time intervals are in the ratio 1:3:5:7:9...1:3:5:7:9... Distance in 1st second (S1S_1) = 12(8)(1)2=4 m\frac{1}{2}(8)(1)^2 = 4 \text{ m}. Distance in 5th second (S5S_5) corresponds to the 5th term (ratio 9). S5=9×S1=9×4=36 mS_5 = 9 \times S_1 = 9 \times 4 = 36 \text{ m}.
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