A body is traveling with a constant acceleration of 10 ms-2. If S1 is the distance traveled in 1st second and S2 is the distance traveled in 2nd second, which of the following shows a correct relation between S1 and S2:

Explanation

The following is the solution: In the 1st second, the body starts from rest (initial velocity, u = 0 m/s) and experiences a constant acceleration (a = 10 m/s²). The distance traveled during this time interval is denoted by S1. Using the equation of motion for constant acceleration: S1 = ut + (1/2)at2 where: u = initial velocity (0 m/s), a = constant acceleration (10 m/s²) and, t = time (1 second) Substituting the values: S1 = 0 + (1/2) x 10 x (1)2 S1 = 0 + 5 S1 = 5 meters So, during the 1st second, the distance traveled (S1) is 5 meters. 2nd second: In the 2nd second, the body continues with the constant acceleration of 10 m/s². The distance traveled during this time interval is denoted by S2. Using the same equation of motion for constant acceleration: S2 = ut + (1/2)at2 where: u = initial velocity (0 m/s) a = constant acceleration (10 m/s²) t = time (2 seconds) Substituting the values: S2 = 0 + (1/2) x 10 x (2)2 S2 = 0 + 20 S2 = 20 meters So, during the 2nd second, the distance traveled (S2) is 20 meters. The correct relation between S1 and S2 is that S2 is four times the value of S1: S2 = 4S1 In this case, S2 (20 meters) is indeed four times the value of S1 (5 meters). Hence option C is correct.