Question

If a body starts from rest and moves with constant acceleration of 2 ms\(^{-2}\), then the distance covered between the time intervals 5 s and 6 s is

• A. 8 m
• B. 15 m
• C. 11 m
• D. 18 m
• E. 6 m

Hint:

The formula \(S_n = u + \frac{a}{2}(2n - 1)\) is a shortcut for finding distance in a single specific second.
It saves you the trouble of calculating two separate displacements and subtracting them.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The distance covered in the \(n^{th}\) second of motion for a body starting with initial velocity \(u\) and constant acceleration \(a\) can be calculated using the specific kinematic equation for displacement in the \(n^{th}\) second.

Step 2: Key Formula or Approach:
\[ S_n = u + \frac{a}{2}(2n - 1) \]

Step 3: Detailed Explanation:
Given:
Initial velocity (\(u\)) = 0 (starts from rest)
Acceleration (\(a\)) = 2 ms\(^{-2}\)
We need the distance between \(t = 5\text{s}\) and \(t = 6\text{s}\). This is exactly the distance covered in the 6th second (\(n = 6\)).
Substituting the values:
\[ S_6 = 0 + \frac{2}{2}(2 \times 6 - 1) \]
\[ S_6 = 1 \times (12 - 1) \]
\[ S_6 = 11 \text{ m} \]
Alternatively:
Calculate \(S\) at \(t=6\) and \(S\) at \(t=5\) using \(S = \frac{1}{2}at^2\):
\(S_6 = \frac{1}{2}(2)(6^2) = 36 \text{ m}\)
\(S_5 = \frac{1}{2}(2)(5^2) = 25 \text{ m}\)
\(\Delta S = 36 - 25 = 11 \text{ m}\).

Step 4: Final Answer:
The distance covered is 11 m.