/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 A particle is dropped from rest ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 24

A particle is dropped from rest from a large height. Assume \(g\) to be constant throughout the motion. The time taken by it to fall through successive distances of \(1 \mathrm{~m}\) each will be (1) All equal, being equal to \(\sqrt{2 / \mathrm{g}}\) second (2) In the ratio of the square roots of the integers \(1,2,3, \ldots .\) (3) In the ratio of the difference in the square roots of the integers, i.e., \(\sqrt{1,(\sqrt{2}-\sqrt{1}),(\sqrt{3}-\sqrt{2}),(\sqrt{4}-\sqrt{3}), \ldots}\) (4) In the ratio of the reciprocals of the square roots of the integers, i.e., \(\frac{1}{\sqrt{1}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}}, \ldots\)

Short Answer

Expert verified
The time intervals are in the ratio of the differences of the square roots of integers.

Step by step solution

01

Understand the Motion Equations

The particle is in free fall, so it starts from rest, meaning its initial velocity is zero. Using the kinematic equation for uniformly accelerated motion: \( s = ut + \frac{1}{2}gt^2 \), where \( s \) is the displacement, \( u = 0 \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( t \) is the time.
02

Calculate First Interval Time

For the first meter of fall, we use \( s = 1 \) m: \[ 1 = \frac{1}{2}gt_1^2 \]\( \Rightarrow t_1 = \sqrt{\frac{2}{g}} \).
03

Generalize for Successive Intervals

For the second meter, calculate \( t_2 \), considering total fall: \[ s = 2 \Rightarrow 2 = \frac{1}{2}gt_2^2 \Rightarrow t_2 = \sqrt{\frac{4}{g}} \]. In general, for the \( n^{th} \) meter, \( t_n = \sqrt{\frac{2n}{g}} \).
04

Calculate Successive Time Differences

The time taken to fall through each meter can be found by: \( \Delta t_n = t_n - t_{n-1} \). So, for each meter: \[ \Delta t_1 = \sqrt{\frac{2}{g}}, \Delta t_2 = \sqrt{\frac{4}{g}} - \sqrt{\frac{2}{g}}, \Delta t_3 = \sqrt{\frac{6}{g}} - \sqrt{\frac{4}{g}} \], etc.
05

Establish Ratio of Successive Times

Calculate the differences in the square roots for each interval: \( t_{12} = \sqrt{2} - \sqrt{1} \), \( t_{23} = \sqrt{3} - \sqrt{2} \), which shows that the times are not equal, but rather in a ratio of the difference between the square roots of successive integers. Hence, the correct ratio is \( \sqrt{1}, (\sqrt{2} - \sqrt{1}), (\sqrt{3} - \sqrt{2}), \ldots \).

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Key Concepts

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

Uniformly Accelerated Motion
Uniformly accelerated motion occurs when a body moves with constant acceleration. In the context of a freely falling object, the acceleration remains constant as it is only influenced by gravity.
  • The object's velocity increases uniformly over time since the force of gravity is unchanging.
  • This means that for equal time intervals, the change in velocity is the same.
The formula used to describe this motion is: \[ s = ut + \frac{1}{2}gt^2 \] where:
  • \(s\) is the displacement, or how far the object has moved from its starting point.
  • \(u\) represents the initial velocity; however, when starting from rest, this becomes zero.
  • \(g\) is the constant acceleration due to gravity.
  • \(t\) is the time for which the object is in motion.
Understanding these elements provides a clarity that aids in calculating the motion of uniformly accelerated objects, such as those in free fall.
Free Fall
Free fall is a special case of uniformly accelerated motion where the only force acting on the object is gravity.
  • Here, the acceleration, denoted as \(g\), is constant because only the gravitational pull is considered.
  • There is no air resistance taken into account in classical free fall problems.
Furthermore, objects begin free fall from rest, meaning their initial velocity \(u\) is zero.
During free fall, without air resistance, all objects accelerate towards the Earth's surface at the same rate, which is a fascinating consequence of gravity's universality. Understanding free fall helps in grasping that the time taken to fall through each successive meter increases due to the increasing velocity.
Acceleration Due to Gravity
The symbol \(g\) represents the acceleration due to gravity. It is approximately \(9.8 \text{ m/s}^2\) on the surface of the Earth.
  • Gravity's pull causes objects in free fall to accelerate at this constant rate, regardless of their mass.
  • It鈥檚 a crucial constant in all equations involving motion near the Earth's surface.
When solving problems involving free fall or other forms of uniformly accelerated motion, this constant helps determine how quickly an object changes velocity.
Taking \(g\) as constant simplifies calculations, allowing students to focus on understanding motion itself without unnecessary complications.
In most educational scenarios, \(g\) is crucial for calculating displacements over time.
Displacement-Time Relation
The displacement-time relation in uniformly accelerated motion provides insights into how far an object will travel over a certain time under constant acceleration.
  • Displacement \(s\) is directly proportional to the square of the time \(t\), when starting from rest.
  • This indicates motion becomes more pronounced as time progresses, due to increasing velocity.
Using the equation \( s = \frac{1}{2}gt^2 \) when \(u = 0\), we see that displacement grows with the square of the time because of consistent gravitational pull.
For our step-by-step solution, this relationship allows calculating the time to fall through successive distances, being proportional to the square roots of consecutive integers
This displays how, as objects continue to fall, their displacement per unit of time increases, setting the foundation for understanding more complex motion dynamics.

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Most popular questions from this chapter

A stone is dropped from a certain height which can reach the ground in \(5 \mathrm{~s}\). It is stopped after \(3 \mathrm{~s}\) of its fall and then it is again released. The total time taken by the stone to reach the ground will be (1) \(6 \mathrm{~s}\) (2) \(6.5 \mathrm{~s}\) (3) \(7 \mathrm{~s}\) (4) \(7.5 \mathrm{~s}\)

Mark the correct statement(s). (1) A particle can have zero displacement and non-zero average velocity. (2) A particle can have zero displacement and non-zero velocity. (3) A particle can have zero acceleration and non-zero velocity. (4) A particle can have zero velocity and non-zero acceleration

The body will speed up if (1) Velocity and acceleration are in the same direction. (2) Velocity and acceleration are in opposite directions. (3) Velocity and acceleration are in perpendicular direction. (4) Velocity and acceleration are acting at acute angle w.r.t. each other.

A parachutist drops first freely from an aeroplane for \(10 \mathrm{~s}\) and then his parachute opens out. Now he descends with a st retardation of \(2.5 \mathrm{~m} \mathrm{~s}^{-2}\). If he bails out of the plane at a theight of \(2495 \mathrm{~m}\) and \(g=10 \mathrm{~ms}^{-2}\), his velocity on reaching the ground will be (1) \(5 \mathrm{~ms}^{-1}\) (2) \(10 \mathrm{~ms}^{-1}\) (3) \(15 \mathrm{~ms}^{-1}\) (4) \(20 \mathrm{~ms}^{-1}\)

A train \(100 \mathrm{~m}\) long travelling at \(40 \mathrm{~ms}^{-1}\) starts overtaking another train \(200 \mathrm{~m}\) long travelling at \(30 \mathrm{~m} \mathrm{~s}^{-1}\). The time taken by the first train to pass the second train completely is (1) \(30 \mathrm{~s}\) (2) \(40 \mathrm{~s}\) (3) \(50 \mathrm{~s}\) (4) \(60 \mathrm{~s}\)
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