A particle starts its motion in a straight line with initial velocity \( 5 \mathrm{~m} . / \mathrm{sec} \). and uniform acceleration \( 2 \mathrm{~m} . / \mathrm{sec}^{2} \). in the same direction of its Initial velocity, then the covered distance during \( 6^{\text {th }}, 7^{\text {th }} \) and \( 8^{\text {th }} \) seconds only \( =\ldots \ldots \ldots \mathrm{m} \). \( \begin{array}{llll}18 & \text { b) } 36 & \text { c) } 54 & \text { d) } 57\end{array} \)

To find the distance covered in the \(6^{\text{th}}, 7^{\text{th}},\) and \(8^{\text{th}}\) seconds, we first use the formula for the distance traveled during any particular second: \[ S_n = u + \frac{1}{2}a(2n - 1) \] where \(u\) is the initial velocity, \(a\) is the acceleration, and \(n\) is the specific second. For \(n = 6\): \[ S_6 = 5 + \frac{1}{2} \times 2 \times (2 \times 6 - 1) = 5 + 1 \times 11 = 16 \, \text{m} \] For \(n = 7\): \[ S_7 = 5 + \frac{1}{2} \times 2 \times (2 \times 7 - 1) = 5 + 1 \times 13 = 18 \, \text{m} \] For \(n = 8\): \[ S_8 = 5 + \frac{1}{2} \times 2 \times (2 \times 8 - 1) = 5 + 1 \times 15 = 20 \, \text{m} \] Now, summing these distances: \[ S = S_6 + S_7 + S_8 = 16 + 18 + 20 = 54 \, \text{m} \] So, the correct answer is \(54 \, \text{m}\). If you want to dive deeper into understanding kinematics, remember that the equations of motion can sometimes feel daunting. Think of it as telling a story where the initial speed, acceleration, and direction are characters that influence the plot. For hands-on practice, consider using real-world objects like a toy car to observe how acceleration affects speed. Measure distances over intervals while giving it a push, and you'll see the principles of kinematics come to life in front of your eyes!