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
2/n + 1/n^2
Answers
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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Option 3) 2n - 1/n^2
Given:
Car starts from rest so the value of u = 0
Calculating the distance covered by the car in the nth second:
Sn = u + a / 2(2n - 1)
= a/2 (2n -1)
Formula used for calculating the distance which is covered in n second:
S = u x n + 1/2 x a x n^2
= 1/2 an^2
(Because we know that u is 0 the car starts from rest)
Now Dividing Sn by S to get the ratio:
Sn / s = a/2(2n - 1)/1/2an^2
= 2n - 1 / n^2
Therefore, the distance covered in the n th second to that covered in n seconds is 2n - 1 / n^2. Option 3) is the correct answer.