Question

a body starts from rest with constant acceleration a, its velocity after n second is v . the displacement of body in last two second is :- 1) 2v (n-1)/ n 2)v (n-1)/n 3)v (n+1)/n 4)2v (n+1)/n

Answer 1

Displacement of the body in ‘n’ sec is, S_n = 0 + ½ an²

Displacement in ‘n – 2’ sec is, S_n-2­ = 0 + ½ a(n - 2)²

So, displacement in the last 2 sec is, S_n – S_n-2 = ½ an² – ½ a(n-2)² = ½ an² – ½ an² + an – 2a = a(n – 2)

Displacement in the n^th second can be found as,

S_n = u + (a/2)(2n – 1)

=> S_n = 0 + (a/2)(2n - 1)

Displacement in the (n-1)^th second can be found as,

S_n-1 = u + (a/2)[2(n-1) - 1]

=> S_n-1 = 0 + (a/2)(2n - 3)

So, displacement in the last two second is,

S_n + S_n-1 = (a/2)(2n - 1) + (a/2)(2n - 3) = (a/2)(2n – 1 + 2n - 3) = (a/2)(4n - 4) = 2a(n - 1)

Now, a = (v - u)/t = v/n

Therefore,

S_n + S_n-1 = (2v/n)(n - 1) = 2v(n - 1)/n