Question

A body starting from rest with an acceleration of \( \frac{5}{4} \, \text{ms}^{-2} \). The distance travelled by the body in the third second is:

• A. \( \frac{15}{8} \, \text{m} \)
• B. \( \frac{25}{8} \, \text{m} \)
• C. \( \frac{25}{4} \, \text{m} \)
• D. \( \frac{12}{7} \, \text{m} \)

Answer 1

The Correct Option is B

We are given the acceleration \( a = \frac{5}{4} \, \text{ms}^{-2} \) and the body starts from rest. We need to find the distance travelled by the body in the third second. The formula for the distance travelled in the \( n^{\text{th}} \) second is given by: \[ S_n = u + \frac{a}{2} (2n - 1), \] where: - \( u \) is the initial velocity (which is 0 since the body starts from rest), - \( a \) is the acceleration, - \( n \) is the time in seconds. We need to find the distance in the third second, so substitute \( n = 3 \) into the equation. The distance travelled in the third second is: \[ S_3 = u + \frac{a}{2} \left( 2 \times 3 - 1 \right). \] Since \( u = 0 \) and \( a = \frac{5}{4} \, \text{ms}^{-2} \), we get: \[ S_3 = \frac{\frac{5}{4}}{2} \times (6 - 1) = \frac{5}{8} \times 5 = \frac{25}{8} \, \text{m}. \] Thus, the distance travelled by the body in the third second is \( \frac{25}{8} \, \text{m} \). Therefore, the correct answer is option (2).

Answer 2

Step 1: Use the formula for distance in the \(n^\text{th}\) second 

The formula for distance travelled in the \( n^\text{th} \) second is: \[ s_n = u + \frac{a}{2}(2n - 1) \] where:

• \( u \) is the initial velocity
• \( a \) is the acceleration
• \( n \) is the second number (here, \( n = 3 \))

 

Step 2: Plug in the values

Given: \[ u = 0, \quad a = \frac{5}{4}, \quad n = 3 \] Then: \[ s_3 = 0 + \frac{1}{2} \cdot \frac{5}{4} \cdot (2 \cdot 3 - 1) = \frac{1}{2} \cdot \frac{5}{4} \cdot 5 = \frac{5 \cdot 5}{8} = \frac{25}{8} \, \text{m} \]

Answer:

\( \boxed{\frac{25}{8} \ \text{m}} \)