/*! 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 28 \(\Lambda\) stone is hanging by ... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\Lambda\) stone is hanging by a thread from a balloon which is moving up with upward acceleration \(a\). Find the total time of motion of the stone when the string is cut after a time \(t\) from starting of the balloon from rest.

Short Answer

Expert verified
The total time of motion of the stone is \(T = t + \frac{2at}{g}\).

Step by step solution

01

Describe the Motion Until the String is Cut

Initially, the balloon is moving with an upward acceleration of \(a\). Let's denote the upward velocity of the stone after time \(t\) as \(v_0\). Since the balloon starts from rest, the velocity \(v_0\) at the time the string is cut can be given by the formula for velocity with constant acceleration: \[ v_0 = a \, t \].
02

Calculate the Distance Covered Before the String is Cut

The distance \(s_0\) the stone has traveled upward before the string is cut can be determined by the formula for distance under constant acceleration: \[ s_0 = \frac{1}{2} a t^2 \].
03

Analyze the Free-Fall Motion After the String is Cut

After the string is cut, the stone is in free fall. The initial velocity for this motion is the upward velocity \(v_0 = a t\). The stone will experience a downward acceleration due to gravity, \(g\).
04

Calculate the Time to Reach Maximum Height

The stone reaches maximum height when its upward velocity becomes zero. Using the equation \(v = u - g t_{up}\), where \(v = 0\) at the maximum height, the time taken to reach this point \(t_{up}\) is \[ t_{up} = \frac{v_0}{g} = \frac{a t}{g} \].
05

Calculate the Time to Fall Back to the Original Height

Once it reaches its maximum height, the stone will fall back to the original height where the string was cut. The time to fall from the highest point back to this height, \(t_{down}\), is the same as the time to reach the maximum height \(t_{up}\): \[ t_{down} = \frac{a t}{g} \].
06

Combine Times for Total Time of Motion

The total time of motion of the stone includes the time until the string is cut plus the time to rise and fall back to the original height. Therefore, the total time, \(T\), is given by: \[ T = t + t_{up} + t_{down} = t + \frac{a t}{g} + \frac{a t}{g} = t + \frac{2at}{g} \].

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Key Concepts

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

Free Fall
The concept of free fall refers to the motion of an object where the only force acting upon it is gravity. This occurs immediately after the stone's string is cut, allowing it to accelerate downwards due to gravity. In free fall, the stone is initially moving upwards due to its velocity at the time of the cut but it soon reaches a point where its upward velocity becomes zero. At this highest point, the stone starts to descend solely under the influence of gravity.

It is important to note that during free fall, air resistance is typically ignored, allowing gravity to act as the lone force. This simplifies the calculations and allows us to focus on the gravitational acceleration, which is approximately 9.8 m/s². This situation exemplifies fundamental principles of physics, particularly motion under constant acceleration.
Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the causes of this motion. For the stone exercise, kinematics helps us understand how the stone moves before and after the string is cut. We calculate the stone's velocity and displacement using kinematic equations.

When the stone is attached to the balloon, its motion is described by constant acceleration upward. Once the string is cut, the free fall motion is described similarly but with gravity. Kinematics allows us to predict various motion aspects like velocity changes and distances covered in the different phases of the stone's journey.
Constant Acceleration
Constant acceleration occurs when an object's velocity changes at a consistent rate over time. In this exercise, when the balloon accelerates upward constantly, the stone moves with the same constant upward acceleration until the string is cut.

The initial velocity of the stone, which is calculated by the equation \(v_0 = a \, t\), is determined by this constant acceleration. After the string is cut, the stone experiences a new constant acceleration downwards due to gravity. Understanding constant acceleration is key to predicting the stone’s velocity and position at various points of time.
Equations of Motion
Equations of motion are vital tools in physics for solving problems involving the motion of objects. These equations help determine velocities and displacements based on factors like initial velocity, time, and constant acceleration.

For instance, the motion of the stone while attached to the balloon involves equations like \(v_0 = a \, t\) for velocity and \(s_0 = \frac{1}{2} a t^2\) for displacement. In the free fall phase, the stone's motion is analyzed using equations that include gravity as the acceleration factor. Equations of motion allow us to calculate the total time of motion by combining the times from each phase: under both upward acceleration and free fall.

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Most popular questions from this chapter

The velocity acquired by a body when it falls through a height \(h\) is \(v\). If it further falls through a very small height \(x\), the increase in velocity is approximately (a) \(\frac{v x}{2 h}\) (b) \(\frac{2 v x}{h}\) (c) \(\frac{2 v}{x h}\) (d) \(0.01 v\)

The velocity of a particle moving in the positive direction o \([x\) -axis varies as \(v-10 \sqrt{x}\). Assuming that at \(t=0\) particle was at \(x=0\). Then (a) The initial velocity of the particle is zero (b) 'The initial velocity of the particle is \(2.5 \mathrm{~m} / \mathrm{s}\) (c) The acceleration of the particle is \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) (d) 'The acceleration of the particle is \(50 \mathrm{~m} / \mathrm{s}^{2}\)

Velocily-time cquation of a particle moving in a straight line is, \(v=\left(10+2 t+3 t^{2}\right) \mathrm{m} / \mathrm{s}\). Find, (a) displacement of particle from the fixed position at time \(t=1 \mathrm{~s}\), il \(\mathrm{it}\) is given that displacement is \(20 \mathrm{~m}\) at time \(t=0\) (b) acecleration-time cquation.

\Lambdat time \(t=0\), a bullet is fired vertically upwards with a speed of \(98 \mathrm{~ms}^{1} . \Lambda\) t time \(t=5 s\) (i.e., 5 seconds later) a second bullet is fired vertically upwards with the same speed. If the air resistance is neglected, which of the following statements will true? (a) 'Ihe two bullets will be at the same height above the ground at \(t=12.5 \mathrm{~s}\). (b) 'lhe two bullets will reach back their starting points at the same time. (c) 'The two bullets will have the same speed at \(t=20 \mathrm{sec}\). (d) 'The two bullets will attain the same maximum height.

A pcrson travelling on a straight linc moves with a uniform velocity \(v_{1}\) for a distance \(x\) and with a unilorm velocity \(v_{2}\) for the next equal distanee The average velocity \(v\) is given by (a) \(v-\frac{v_{1}+v_{2}}{2}\) (b) \(v-\sqrt{v_{1} v_{2}}\) (c) \(\frac{2}{v}-\frac{1}{v_{1}}+\frac{1}{v_{2}}\) (d) \(\frac{1}{v}-\frac{1}{v_{1}}+\frac{1}{v_{2}}\)

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