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 solve this problem, we need to understand the motion of a body under constant acceleration. The given information tells us that the body starts from rest and covers displacements \( S_1 \) in \((p - 1)\) seconds and \( S_2 \) in \( p \) seconds. We want to find the time in which the displacement \( S_1 + S_2 \) is covered. 

Let's break down the solution with step-by-step reasoning:

1. Since the body starts from rest with constant acceleration, the displacement covered in time \( t \) is given by the formula: \(S = \frac{1}{2} a t^2\), where \( a \) is the constant acceleration.
2. For the first \((p - 1)\) seconds, the displacement \( S_1 \) is: \(S_1 = \frac{1}{2} a (p-1)^2\).
3. For the first \( p \) seconds, the displacement \( S_2 \) is: \(S_2 = \frac{1}{2} a p^2\).
4. The displacement in the \( p\)th second, which is the difference \( S_2 - S_1 \), can be calculated as: \(\Delta S = \frac{1}{2} a (p^2 - (p-1)^2)\).
5. Simplify \( p^2 - (p-1)^2 \): \(p^2 - (p-1)^2 = p^2 - (p^2 - 2p + 1) = 2p - 1\).
6. Substituting this back gives us: \(\Delta S = \frac{1}{2} a (2p - 1)\).
7. We are asked to find the time in which the total displacement \( S_1 + S_2 \) occurs. The total displacement is: \(S_{\text{total}} = S_1 + \Delta S = \frac{1}{2} a p^2\).
8. To find the time needed to cover the displacement \( S_{\text{total}} \), we equate it to the general formula: \(\frac{1}{2} a t^2 = \frac{1}{2} a p^2\).
9. Solving for \( t \) gives: \(t = \sqrt{p^2 - (p - 1)^2 + 2p - 1} = \sqrt{2p^2 - 2p + 1}\).

Thus, the time in which the displacement \( S_1 + S_2 \) is covered is the option: \(\sqrt{2p^2 - 2p + 1} \, s\).

Therefore, the correct option is:

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

 

Answer 2

Step 1: Calculate \(S_1\) in the First \((p - 1)\) Seconds

Since the body starts from rest, using the formula \(S = \frac{1}{2}at^2\),

\[ S_1 = \frac{1}{2}a(p - 1)^2 \]

Step 2: Calculate \(S_2\) in the First \(p\) Seconds

Using the same formula for \(S_2\),

\[ S_2 = \frac{1}{2}ap^2 \]

Step 3: Total Displacement \(S_1 + S_2\)

If \(S_1 + S_2\) represents the displacement in time \(t\), then:

\[ S_1 + S_2 = \frac{1}{2}at^2 \]

Substitute \(S_1\) and \(S_2\) values:

\[ \frac{1}{2}a(p - 1)^2 + \frac{1}{2}ap^2 = \frac{1}{2}at^2 \]

Simplify by canceling \(\frac{1}{2}a\):

\[ (p - 1)^2 + p^2 = t^2 \]

Step 4: Solve for \(t\)

\[ t = \sqrt{2p^2 - 2p + 1} \]

So, the correct answer is: \(\sqrt{2p^2 - 2p + 1} \, s\)