/*! 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 43 Concept Simulation 2.3 at offers... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Concept Simulation 2.3 at offers a useful review of the concepts central to this problem. An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of \(+15 \mathrm{~m} / \mathrm{s}\) and measures a time of \(20.0 \mathrm{~s}\) before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet?

Short Answer

Expert verified
The acceleration due to gravity on the planet is \(-1.5 \, \text{m/s}^2\) downward.

Step by step solution

01

Understand the Problem

The astronaut throws a rock upwards and it returns to his hand, taking a total of 20 seconds. We need to find the acceleration due to gravity on this distant planet, using the given initial velocity and time.
02

Break Down the Motion

Since the rock returns to the astronaut's hand, we can split the motion into two parts: the upward journey and the downward journey. Each of these journeys takes 10 seconds (since the total round trip time is 20 seconds).
03

Apply the Kinematic Equation

Use the equation for velocity under constant acceleration: \[ v = u + at \]where \(v\) is the final velocity (0 m/s at the peak), \(u = 15 \text{ m/s}\) is the initial velocity, \(a\) is the acceleration, and \(t = 10 \text{ s}\) is the time taken to reach the peak. Solve for \(a\).
04

Solve for the Acceleration

Rearrange the equation: \[ 0 = 15 + a(10) \]Solving for \(a\) gives:\[ a = \frac{-15}{10} = -1.5 \, \text{m/s}^2 \]The negative sign indicates that the acceleration is directed downward.

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 91Ó°ÊÓ!

Key Concepts

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

Understanding Kinematics
Kinematics is a branch of physics that describes the motion of objects without considering the forces causing the motion. This concept is fundamental in physics as it allows us to analyze how objects move through space and time. In the exercise, we dealt with a rock being thrown upwards on a distant planet.
Kinematics often involves the use of various kinematic equations to predict and describe the motion of objects. A key element of kinematics is understanding that motion can happen in different phases, such as the upward and downward phases of the rock's journey, as described in the exercise.
In real-world scenarios, kinematics can help us:
  • Calculate distances traveled by objects over time.
  • Determine the velocity or speed of objects at specific points in their motion.
  • Understand acceleration - how the velocity of an object changes.
This discipline is crucial when analyzing movement in planes, cars, or even celestial bodies like planets.
Exploring Constant Acceleration
Constant acceleration refers to a situation where the rate of change of velocity remains the same throughout the motion. In the given problem, we're looking at the rock experiencing the same gravitational acceleration as it travels up and down.
Kinematic equations are typically used to solve problems involving constant acceleration. These equations provide relationships between initial velocity, final velocity, acceleration, time, and displacement. In our problem, the kinematic equation used is:\[ v = u + at \]where \(v\) is the final velocity, \(u\) the initial velocity, \(a\) the acceleration, and \(t\) the time taken. This equation is particularly useful when dealing with vertical motion under the influence of gravity, which is an example of constant acceleration.
Constant acceleration is a common assumption in physics because many systems, like gravity here on Earth, act with a practically constant force on objects allowing us to predict the motion precisely.
Explaining Motion in One Dimension
Motion in one dimension refers to movement along a straight line, which simplifies the analysis of an object's path. For instance, in the exercise, the rock was only moving up and down along a vertical path, making it a one-dimensional motion problem.
When analyzing one-dimensional motion, we only consider one spatial direction. This reduction in complexity allows us to focus on how the object's position changes over time due to factors like speed, velocity, time, and acceleration.
Key aspects of one-dimensional motion will include:
  • Initial and final positions, which can be used to find displacement.
  • Initial and final velocities, important for determining how speed changes.
  • Constant or changing acceleration, influencing the velocity and direction of an object.
Motion in one dimension is foundational in physics because it underpins more complex movements that occur in two or three dimensions.

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

An 18 -year-old runner can complete a \(10.0\) -km course with an average speed of \(4.39 \mathrm{~m} / \mathrm{s}\). A 50 -year-old runner can cover the same distance with an average speed of \(4.27 \mathrm{~m} / \mathrm{s}\). How much later (in seconds) should the younger runner start in order to finish the course at the same time as the older runner?

Review Conceptual Example 7 as background for this problem. A car is traveling to the left, which is the negative direction. The direction of travel remains the same throughout this problem. The car's initial speed is \(27.0 \mathrm{~m} / \mathrm{s},\) and during a 5.0 -s interval, it changes to a final speed of (a) \(29.0 \mathrm{~m} / \mathrm{s}\) and ( b) \(23.0 \mathrm{~m} / \mathrm{s}\). In each case, find the acceleration (magnitude and algebraic sign) and state whether or not the car is decelerating.

From her bedroom window a girl drops a water-filled balloon to the ground, \(6.0 \mathrm{~m}\) below. If the balloon is released from rest, how long is it in the air?

ssm The greatest height reported for a jump into an airbag is \(99.4 \mathrm{~m}\) by stuntman Dan Koko. In 1948 he jumped from rest from the top of the Vegas World Hotel and Casino. He struck the airbag at a speed of \(39 \mathrm{~m} / \mathrm{s}(88 \mathrm{mi} / \mathrm{h})\). To assess the effects of air resistance, determine how fast he would have been traveling on impact had air resistance been absent.

(a) What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) when going down a slope for \(5.0 \mathrm{~s} ?\) (b) How far does the skier travel in this time?

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.