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Chapter 3: Problem 35

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

Expert verified
The ratio is option (a) \( \frac{2}{n} - \frac{1}{n^2} \).

Step by step solution

01

Understanding the Key Concepts

The body starts from rest, meaning initial velocity is zero, and it moves with constant acceleration. Using the relation between acceleration, velocity, and distance will be key in solving this problem.
02

Using the Equation for Distance Covered

The distance covered in the nth second can be obtained using the formula for distance in nth second: \( S_n = u + \frac{a}{2}(2n-1) \) where \( u = 0 \). This simplifies to \( S_n = \frac{a}{2}(2n-1) \).
03

Total Distance in n Seconds

The total distance covered in \( n \) seconds can be calculated using the formula \( S = \frac{1}{2}a n^2 \) since the body started from rest.
04

Calculating the Ratio

We want the ratio of the distance covered in the nth second to the total distance in \( n \) seconds. Using the values obtained, the ratio is \( \frac{\frac{a}{2}(2n-1)}{\frac{1}{2}an^2} = \frac{2(2n-1)}{n^2} \). Simplifying this ratio gives \( \frac{2}{n} - \frac{1}{n^2} \).
05

Choosing the Correct Option

Compare the simplified ratio with the given options. The correct answer is option (a) \( \frac{2}{n} - \frac{1}{n^2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Acceleration
In the realm of kinematics, understanding motion is crucial. One fundamental idea is constant acceleration. When an object moves with constant acceleration, its velocity changes at a steady rate. This means that the change in velocity occurs evenly over time. Constant acceleration plays a significant role in many real-world scenarios, from cars accelerating on a highway to objects falling under gravity.

To comprehend constant acceleration more fully, consider this:
  • The initial velocity (\( u \)) is the velocity the object starts with. If an object starts from rest, \( u = 0 \).
  • Acceleration (\( a \)) is the rate of change of velocity. During constant acceleration, \( a \) remains unchanged.
  • This steady acceleration influences how quickly and in what manner the velocity and position of the object change.
These principles form the foundation of many equations of motion, making constant acceleration a pivotal concept to understand.
Distance Calculation
When calculating distances in kinematics, it's essential to use the right formulas. Distance calculation can become tricky, especially when dealing with constant acceleration.

In this exercise, we calculated the distance covered in the \( n \) th second and the total distance covered in \( n \) seconds:
  • To find the distance covered in a specific second, use the formula \( S_n = u + \frac{a}{2}(2n-1) \). Here, because the object starts from rest, \( u = 0 \), simplifying the formula to \( S_n = \frac{a}{2}(2n-1) \).
  • The total distance after \( n \) seconds is found with \( S = \frac{1}{2}an^2 \). This formula helps calculate the accumulated distance as the body accelerates constantly over time.
Understanding these formulas allows you to solve not just theoretical exercises but also practical problems involving moving objects.
Equations of Motion
The equations of motion are essential in solving various problems in physics, particularly those involving constant acceleration. These equations give us the tools needed to calculate velocity, acceleration, time, and distance.

In exercises involving motion, these key equations are predominantly used:
  • Final velocity (\( v \)) can be calculated when initial velocity (\( u \)), acceleration (\( a \)), and time (\( t \)) are known: \( v = u + at \).
  • Total distance (\( S \)) for an object from rest can be determined using: \( S = \frac{1}{2} a n^2 \).
  • The specific distance covered in the \( n \) th second can be found: \( S_n = \frac{a}{2}(2n-1) \).
These formulas help break down complex motion problems into simple calculations. By understanding and using these equations, students can predict and analyze motion very precisely.

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Most popular questions from this chapter

A particle starts with a velocity of \(2 \mathrm{~m} / \mathrm{s}\) and moves in a straight line with a retardation of \(0.1 \mathrm{~m} / \mathrm{s}^{2}\). The time that it takes to describe \(15 \mathrm{~m}\) is : (a) \(10 \mathrm{~s}\) in its backward journey (b) \(30 \mathrm{~s}\) in its forward journey (c) \(10 \mathrm{~s}\) in its forward journey (d) \(30 \mathrm{~s}\) in its backward journey (e) both (b) and (c) are correct

Rain water is falling vertically downward with velocity \(v\). When velocity of wind is \(u\) in horizontal direction, water is collected at the rate of \(\mathrm{R} \mathrm{m}^{3} / \mathrm{s}\). When velocity of wind becomes \(2 u\) in horizontal direction, the rate of collection of water in vessel is: (a) \(R\) (b) \(\frac{R}{2}\) (c) \(2 R\) (d) \(\frac{R \sqrt{4 u^{2}+v^{2}}}{\sqrt{u^{2}+v^{2}}}\)

When acceleration of a particle is \(a=f(t)\), then: (a) the velocity, starting from rest is \(\int_{n}^{i} f(t) d t\) (b) velocity may be constant (c) the velocity must not be function of time (d) the speed may be constant with respect to time

A cat wants to catch a rat. The cat follows the path whose equation is \(x+y=0 .\) But rat follows the path whose equation is \(x^{2}+y^{2}=4\). The co- ordinates of possible points of catching the rat are: (a) \((\sqrt{2}, \sqrt{2})\) (b) \((-\sqrt{2}, \sqrt{2})\) (c) \((\sqrt{2}, \sqrt{3})\) (d) \((0,0)\)

The locus of foci of all parabolas described by the particles projected simultaneously from the same point with equal velocities but in different directions is a : (a) circle (b) parabola (c) ellipse (d) hyperbola
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