Question

A car starts from rest and moves with constant acceleration. The ratio of the distance covered in the nth second to that covered in n seconds is:

1) 2/n^2 - 1/n
2) 2/n^2 + 1/n
3) 2/n - 1/n^2
4) 2/n + 1/n^2

Answer 1

Values given

The car starts from rest .

u = 0

Velocity in n - 1  seconds

Till the n  - 1 th second , the time is n - 1

let velocity be v

v = u + a t

 = 0 + a ( n - 1 )

v = a ( n - 1 )

Velocity in the n seconds

v = u + a t

 ==> 0 + a × n

 ==> a n

Average velocity in n seconds :

Average velocity = [ a n + a ( n - 1 ) ] / 2

                             ==> [ an + an - a ] / 2

                             ==> [ 2 an - a ] / 2 .......................(1)

Distance in n seconds

Average velocity = total distance / time for  that distance

==> distance = average velocity × time

                         for n seconds

==> distance = [ 2 an - a ] / 2 ×1

                      ==> [ 2 an - a ] / 2 .......................(2)

                          for the n th second

==> distance = S = ut + 1/2 at²

==> distance = S = 0 + 1/2 a ×n²

                            = 1/2 a n² ............................(3)

Ratio of distance in nth second to n seconds is :

                            Divide (2) and (3)

( [ 2 an - a ] / 2 ) = 1/2 a n²

Cancelling 2 we get :

==> [ 2 an - a ] / a n²

==> a [ 2 n - 1 ] / an²

==> [ 2 n - 1 ] / n²

The correct answer is hence  OPTION (3) [ ( 2 n - 1 ) / n² ]

Hope it helps you :)

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Answer 2

Answer:

option 3

Explanation:

Velocity at the end of (n-1) sec = U +at = 0+ a*(n-1)

Velocity at the end of n th sec = a*(n-1) + a*1 = na

Average velocity in the nth sec = (a*(n-1) + n*a) /2 = (a/2)(2n-1)

Distance covered in the n th sec =

Average velocity in the nth sec* time( =1) = (a/2)(2n-1)

Distance covered in n sec = 1/2 * a * t² = a/2 * n²

The desired ratio = (2n - 1) / n²