Chapter 2: Problem 8
A body starts from rest and moves with a constant acceleration. The ratio of distance covered in the \(n\)th second to the distance covered in \(n\) second is (a) \(\frac{2}{n}-\frac{1}{n^{2}}\) (b) \(\frac{1}{n^{2}}-\frac{1}{n}\) (c) \(\frac{2}{n^{2}}-\frac{1}{n}\) (d) \(\frac{2}{n}+\frac{1}{n^{2}}\)
Short Answer
Step by step solution
Understanding the Problem
Formula for Distance in nth Second
Total Distance in n Seconds
Ratio of Distances
Simplifying the Ratio Expression
Conclusion and Verification
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Distance-Time Relationship
- \(S\) is the total distance covered,
- \(u\) is the initial velocity,
- \(a\) is the constant acceleration,
- \(t\) is the time elapsed.
This equation tells us that the distance covered by the object is proportional to the square of the time elapsed, which is a hallmark of motion under constant acceleration. This quadratic relationship implies that as time increases, the distance the object covers grows faster, making it an essential concept for solving various motion problems.
Nth Second Formula
For instance, if you want to know how much ground the object covers specifically in the 5th second, this formula helps us find that specific value, distinguishing it from the total ground up to that point. It's an essential tool for deeper insights into kinematics, clarifying how acceleration affects motion over discrete time intervals.
Kinematic Equations
- \(v = u + at\)
- \(S = ut + \frac{1}{2}at^2\)
- \(v^2 = u^2 + 2aS\)
- \(S = \frac{(u + v)}{2}t\)
- Velocity changes due to constant acceleration,
- Distance covered over time with constant acceleration,
- Relating initial velocity, final velocity, and distance covered,
- The average velocity if acceleration is constant.