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 determine the distance a particle travels during its 5th second of motion, given it begins from rest and maintains a constant acceleration of 4 m/s\(^2\), employ the formula for displacement in the nth second:

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

The variables are defined as follows:

• \(s_n\) represents the distance covered in the nth second.
• \(u\) is the initial velocity, which is 0 m/s since the particle starts from rest.
• \(a\) denotes the acceleration, given as 4 m/s\(^2\).
• \(n\) signifies the specific second being analyzed, in this instance, 5.

Upon substituting the provided values into the formula:

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

\(s_5 = 2 \times 9\)

\(s_5 = 18 \, m\)

Consequently, the particle covers a distance of 18 meters in the 5th second.