Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

An elevator ascends with an upward acceleration of \(4.0-\mathrm{ft} / \mathrm{s}^{2}\). At the instant its upward speed is \(8.0 \mathrm{ft} / \mathrm{s}\), a loose bolt drops from the ceiling of the elevator \(9.0 \mathrm{ft}\) from the floor. Calculate (a) the time of flight of the bolt from ceiling to floor and \((b)\) the distance it has fallen relative to the elevator shaft.

Short Answer

Expert verified
The time of flight of the bolt from the ceiling to the floor is the smaller of the two roots obtained from step 3, and the distance the bolt has fallen relative to the elevator shaft is obtained in step 4.

Step by step solution

01

Determine the initial speed of the bolt

The bolt has the same initial upward speed as the elevator when it is dropped, which is \(8.0 \, ft/s\).
02

Calculate the time of flight of the bolt

The bolt is under the effect of two accelerations: the acceleration due to gravity, which is \(32.2 \, ft/s^2\) in the downward direction, and the elevator's acceleration of \(4.0 \, ft/s^2\) in the upward direction. Consequently, the net acceleration is \(32.2 - 4.0 = 28.2 \, ft/s^2\) in the downward direction. Use the formula: \[ h = v_0t + 0.5at^2 \] where \(h\) is the height, \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time to solve for \(t\): \[ 9.0 = 8.0t + 0.5(-28.2)t^2 \] Solve the above equation for \(t\) to find both roots.
03

Select the correct time root

Select the smaller positive root of \(t\) as the falling object cannot take negative time or a longer time which is the larger root to fall.
04

Calculate the relative distance from elevator shaft

The distance the bolt falls relative to the elevator shaft can be calculated by considering the motion of the bolt from the point of view of an observer in the elevator. From the frame of reference of the elevator, the bolt has an initial velocity of \(8.0 \, ft/s\) and a constant acceleration of \(32.2 \, ft/s^2\) downward (the acceleration due to gravity). So the net distance fallen relative can be calculated using \(h = v_0t + 0.5at^2\) where \(h\) is the distance, \(v_0\) is the initial velocity which is \(8.0 \, ft/s\), \(a\) is the acceleration, which is \(32.2 \, ft/s^2\), and \(t\) is the time obtained in step 3. Calculate \(h\) by plugging in these values.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Newton's Laws of Motion
Newton's Laws of Motion are fundamental principles in classical mechanics that describe the relationship between a body and the forces acting upon it. In the problem with the elevator, Newton's Second Law is especially relevant.
By understanding that force equals mass times acceleration (\( F = ma \)), we can interpret how different forces affect the motion of objects. In the case of the falling bolt, two main forces are in play.
  • First, the upward force from the elevator's acceleration.
  • Second, the gravitational force pulling it downwards.

Because of these opposing forces, the net acceleration is adjusted and primarily determines how quickly the bolt moves relative to its starting point and the elevator. The net effect of these forces provides the bolt with a downward acceleration, leading to its descent to the floor of the elevator.
Kinematics
Kinematics is another crucial aspect of this problem. This branch of mechanics is concerned with the motion of objects without considering the forces that cause them. It provides the mathematical framework to predict the path an object will take.

In our exercise, we utilize kinematic equations to determine the time it takes for the bolt to hit the floor. The kinematic equation \[ h = v_0t + 0.5at^2 \] enables us to solve for the unknown time (\( t \)). Here:
  • \( h \) is the distance the bolt falls, which is 9.0 ft.
  • \( v_0 \) is the initial speed of the bolt, taken from the elevator, at 8.0 ft/s.
  • \( a \) is the net acceleration, obtained by subtracting the elevator's upward acceleration from gravity.
The method to solve for the two potential roots of time emerges from applying this formula, ensuring we select the physically relevant positive time interval.
Relative Motion
Relative motion examines how different observers perceive movement differently depending on their own motion. This perspective becomes clear with the bolt: it falls in the context of the moving elevator.

From the standpoint of someone inside the elevator, the bolt appears to drop straight down to the floor. As the elevator ascends with acceleration, the apparent trajectory of the bolt is a straightforward drop through its initial 9 ft height.

However, if standing still within the elevator shaft, the bolt seems to follow a more complex path, accelerating downward faster than someone moving upwards with the elevator experiences. The equation \[ h = v_0t + 0.5at^2 \] is applied again, but now using \( a = 32.2 \, ft/s^2 \) (acceleration due to gravity), showing the distance fallen relative to the elevator shaft.
  • This concept highlights how the same bolt drop is viewed differently depending on the observer and their frame of reference.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a baseball game, a batter hits the ball at a height of \(4.60 \mathrm{ft}\) above the ground so that its angle of projection is \(52.0^{\circ}\) to the horizontal. The ball lands in the grandstand, \(39.0 \mathrm{ft}\) up from the bottom; see Fig. 4-38. The grandstand seats slope upward at \(28.0^{\circ}\) with the bottom seats \(358 \mathrm{ft}\) from home plate. Calculate the speed with which the ball left the bat. (Ignore air resistance.)

A light plane attains an air speed of \(480 \mathrm{~km} / \mathrm{h}\). The pilot sets out for a destination \(810 \mathrm{~km}\) to the north but discovers that the plane must be headed \(21^{\circ}\) east of north to fly there directly. The plane arrives in \(1.9 \mathrm{~h}\). What was the vector wind velocity?

Certain neutron stars (extremely dense stars) are believed to be rotating at about 1 rev/s. If such a star has a radius of \(20 \mathrm{~km}\) (a typical value), \((a)\) what is the speed of a point on the equator of the star, and \((b)\) what is the centripetal acceleration of this point?

In a cathode-ray tube, a beam of electrons is projected horizontally with a speed of \(9.6 \times 10^{8} \mathrm{~cm} / \mathrm{s}\) into the region between a pair of horizontal plates \(2.3 \mathrm{~cm}\) long. An electric field between the plates causes a constant downward acceleration of the electrons of magnitude \(9.4 \times 10^{16} \mathrm{~cm} / \mathrm{s}^{2} .\) Find \((a)\) the time required for the electrons to pass through the plates, \((b)\) the vertical displacement of the beam in passing through the plates, and (c) the horizontal and vertical components of the velocity of the beam as it emerges from the plates.

A train travels due south at \(28 \mathrm{~m} / \mathrm{s}\) (relative to the ground) in rain that is blown to the south by the wind. The path of each raindrop makes an angle of \(64^{\circ}\) with the vertical, as measured by an observer stationary on the Earth. An observer on the train, however, sees perfectly vertical tracks of rain on the windowpane. Determine the speed of the drops relative to the Earth.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free