A particle starts from rest and moves with constant acceleration for a total time t seconds. The distance covered by the particle in the last 2 seconds of its motion is exactly equal to the distance covered in the first 4 seconds. The ratio of the total distance covered during the entire motion to the distance covered in the first second is ______.

Step-by-step solution

Let the constant acceleration be a . The distance covered in the first 4 seconds is S_ first 4 = 1 2 a(4)^2 = 8a . The distance covered in the last 2 seconds of the total time t is the total distance minus the distance covered in (t-2) seconds: S_ last 2 = 1 2 at^2 - 1 2 a(t-2)^2 S_ last 2 = 1 2 a[t^2 - (t^2 - 4t + 4)] = 1 2 a(4t - 4) = 2a(t - 1) Given that these two distances are equal: 2a(t - 1) = 8a t - 1 = 4 t = 5 s. The total distance covered during the entire motion ( t=5 s) is D = 1 2 a(5)^2 = 25a 2 . The di