Question

A body starts from rest with uniform acceleration and its velocity at a time of \( n \) seconds is \( v \). The total displacement of the body in the \( n \)-th and \( (n - 1) \)-th seconds of its motion is

• A. \( \frac{v(n+1)}{n} \)
• B. \( \frac{2v(n+1)}{n} \)
• C. \( \frac{2v(n-1)}{n} \)
• D. \( \frac{v(n-1)}{n} \)

Hint:

For displacement in the \( m \)-th second under uniform acceleration, use \( S_m = u + \frac{1}{2} a (2m - 1) \). Relate velocity and acceleration via \( v = u + a t \).

Answer 1

The Correct Option is C

The body starts from rest (\( u = 0 \)) with uniform acceleration \( a \). The velocity at \( t = n \) seconds is \( v = u + a n = a n \), so: \[ a = \frac{v}{n} \] The displacement in the \( m \)-th second is given by: \[ S_m = u + \frac{1}{2} a (2m - 1) \] For the \( n \)-th second (\( m = n \)): \[ S_n = 0 + \frac{1}{2} \cdot \frac{v}{n} (2n - 1) = \frac{v (2n - 1)}{2n} \] For the \( (n - 1) \)-th second (\( m = n - 1 \)): \[ S_{n-1} = 0 + \frac{1}{2} \cdot \frac{v}{n} (2(n - 1) - 1) = \frac{v (2n - 3)}{2n} \] Total displacement: \[ S_n + S_{n-1} = \frac{v (2n - 1)}{2n} + \frac{v (2n - 3)}{2n} = \frac{v (2n - 1 + 2n - 3)}{2n} = \frac{v (4n - 4)}{2n} = \frac{2v (n - 1)}{n} \] Option (3) is correct. Options (1), (2), and (4) do not match.