Question

A body starts moving from rest with constant acceleration and covers displacement \(S_1\) in the first \((p - 1)\) seconds and \(S_2\) in the first \(p\) seconds. The displacement \(S_1 + S_2\) will be made in time:

• A. \(\sqrt{2p^2 - 2p + 1} \, s\)
• B. \((2p + 1) \, s\)
• C. \((2p - 1) \, s\)
• D. \((2p^2 - 2p + 1) \, s\)

Answer 1

The Correct Option is A

To address this problem, an understanding of motion under constant acceleration is required. The problem states that a body begins from rest and travels distances \( S_1 \) over \((p - 1)\) seconds and \( S_2 \) over \( p \) seconds. The objective is to determine the time required to cover the combined displacement \( S_1 + S_2 \).

The solution is presented step-by-step:

1. Given the body starts from rest with constant acceleration \( a \), the displacement \( S \) covered in time \( t \) is calculated using the formula: \(S = \frac{1}{2} a t^2\).
2. The displacement \( S_1 \) in the initial \((p - 1)\) seconds is: \(S_1 = \frac{1}{2} a (p-1)^2\).
3. The total displacement \( S_2 \) after \( p \) seconds is: \(S_2 = \frac{1}{2} a p^2\).
4. The displacement covered specifically within the \( p \)th second, represented as \( S_2 - S_1 \), is: \(\Delta S = \frac{1}{2} a (p^2 - (p-1)^2)\).
5. The term \( p^2 - (p-1)^2 \) simplifies to: \(p^2 - (p^2 - 2p + 1) = 2p - 1\).
6. Substituting this simplified term yields: \(\Delta S = \frac{1}{2} a (2p - 1)\).
7. The total displacement to be considered is \( S_1 + S_2 \). However, the problem seems to re-interpret \(S_2\) as the displacement in the \(p^{th}\) second. Assuming \(S_2\) refers to the displacement in the \(p^{th}\) second (i.e., \(S_2 = \Delta S\)), the total displacement is \(S_{\text{total}} = S_1 + S_2 = \frac{1}{2} a (p-1)^2 + \frac{1}{2} a (2p - 1) = \frac{1}{2} a (p^2 - 2p + 1 + 2p - 1) = \frac{1}{2} a p^2\).
8. To ascertain the time \( t \) required for this total displacement, we equate it to the general displacement formula: \(\frac{1}{2} a t^2 = \frac{1}{2} a p^2\).
9. Solving for \( t \) yields: \(t = p\).
10. Re-evaluating the problem statement, it is more likely that \(S_1\) is the displacement in \((p-1)\) seconds and \(S_2\) is the displacement in the \(p^{th}\) second. In this case, the total displacement is \(S_{\text{total}} = S_1 + S_2 = \frac{1}{2} a (p-1)^2 + \frac{1}{2} a (2p - 1) = \frac{1}{2} a p^2\). The time to cover this displacement is \(t=p\) seconds.
11. If \( S_1 \) is the displacement in \((p - 1)\) seconds and \( S_2 \) is the displacement in \(p\) seconds, then the displacement in the \(p^{th}\) second is \(S_2 - S_1\). The problem asks for the time to cover \( S_1 + S_2 \). This interpretation is problematic as \(S_2\) already includes the displacement \(S_1\).
12. Let's assume the question meant: \(S_1\) is the displacement in \((p-1)\) seconds, and \(S_2\) is the displacement in the \(p^{th}\) second. Then the total displacement covered up to time \(p\) is \(S_1 + S_2 = \frac{1}{2} a (p-1)^2 + \frac{1}{2} a (2p-1) = \frac{1}{2} a p^2\). The time to cover this is \(p\) seconds.
13. Let's reconsider the original interpretation: \( S_1 \) in \((p - 1)\) seconds and \( S_2 \) in \( p \) seconds. The displacement in the \( p^{th} \) second is \( \Delta S = S_2 - S_1 \). We need to find the time to cover \( S_1 + S_2 \). The total displacement up to time \( p \) is \( S_2 \). This interpretation is also confusing.
14. Assuming the question intends to find the time to cover a displacement equal to the sum of the displacement up to \((p-1)\) seconds and the displacement in the \(p^{th}\) second. This sum is \(S_1 + (S_2 - S_1) = S_2 = \frac{1}{2} a p^2\). The time to cover this is \(p\) seconds.
15. Another interpretation: The problem asks for the time to cover a displacement that is the sum of the displacement in \((p-1)\) seconds and the displacement up to \(p\) seconds. This would be \( S_1 + S_2 = \frac{1}{2} a (p-1)^2 + \frac{1}{2} a p^2 \). The time to cover this is \( t \) such that \( \frac{1}{2} a t^2 = \frac{1}{2} a ((p-1)^2 + p^2) \), so \( t = \sqrt{p^2 + (p-1)^2} = \sqrt{2p^2 - 2p + 1} \).
16. This last interpretation aligns with the provided solution structure. Let \(S(t)\) be the displacement at time \(t\). We are given \(S(p-1) = S_1\) and \(S(p) = S_2\). We want to find the time \( T \) such that \( S(T) = S_1 + S_2 \).
17. \( S_1 = \frac{1}{2} a (p-1)^2 \)
18. \( S_2 = \frac{1}{2} a p^2 \)
19. We need to find \( T \) such that \( S(T) = \frac{1}{2} a T^2 = S_1 + S_2 = \frac{1}{2} a (p-1)^2 + \frac{1}{2} a p^2 \).
20. \( \frac{1}{2} a T^2 = \frac{1}{2} a [(p-1)^2 + p^2] \)
21. \( T^2 = (p-1)^2 + p^2 = p^2 - 2p + 1 + p^2 = 2p^2 - 2p + 1 \)
22. \( T = \sqrt{2p^2 - 2p + 1} \)

The time required to cover the displacement \( S_1 + S_2 \) is therefore: \( \sqrt{2p^2 - 2p + 1} \) seconds.

The correct option is:

\(\sqrt{2p^2 - 2p + 1} \, s\)