Question

A particle starts from rest and accelerates uniformly at 4 m/s\(^2\). What is the distance covered in the 5th second?

• A. 36 m
• B. 18 m
• C. 44 m
• D. 20 m

Hint:

Formula: For uniformly accelerated motion, use \( s_n = u + \frac{a}{2}(2n - 1) \) to find distance in the \(n^{\text{th}}\) second.

Answer 1

The Correct Option is B

To find the distance covered by a particle in the 5th second, when it starts from rest and accelerates uniformly at 4 m/s\(^2\), we can use the formula for distance covered in the nth second:

\(s_n = u + \frac{a}{2} \times (2n-1)\)

Where:

• \(s_n\) is the distance covered in the nth second
• \(u\) is the initial velocity (0 m/s as the particle starts from rest)
• \(a\) is the acceleration (4 m/s\(^2\))
• \(n\) is the specific second of interest (5 in this case)

Substituting the given values into the formula:

\(s_5 = 0 + \frac{4}{2} \times (2 \times 5 - 1)\)

\(s_5 = 2 \times 9\)

\(s_5 = 18 \, m\)

Therefore, the distance covered in the 5th second is 18 meters.

Answer 2

• Step 1: Use the formula for distance in the \(n^{\text{th}}\) second:
  \[ s_n = u + \frac{a}{2}(2n - 1) \] where \( u = 0 \), \( a = 4 \, \text{m/s}^2 \), \( n = 5 \)
• Step 2: Substitute values.
  \[ s_5 = 0 + \frac{4}{2}(2 \cdot 5 - 1) = 2 \cdot 9 = \boxed{18 \text{ m}} \] Correct Answer: (A) 18 m