/*! 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 73 You are given the kinematic equa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 73

You are given the kinematic equation or equations that are used to solve a problem. For each of these, you are to: a. Write a realistic problem for which this is the correct equation(s). Be sure that the answer your problem requests is consistent with the equation(s) given. b. Draw the pictorial representation for your problem. c. Finish the solution of the problem. $$(10 \mathrm{m} / \mathrm{s})^{2}=v_{0 y}^{2}-2\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(10 \mathrm{m}-0 \mathrm{m})$$

Short Answer

Expert verified
It seems like the equation given is incorrect if we follow the usual behavior of a falling object. The right-hand side of the equation ends up as a negative value, which is not possible as velocity squared cannot be negative.

Step by step solution

01

Formulate Real World Scenario

A stone is dropped from a height of 10 meters. The velocity when it hits the ground is required to find real-world physics.
02

Pictorial Representation

The picture would show a stone falling from a height of 10 meters towards the ground. As it falls, the stone accelerates due to the force of gravity.
03

Calculate Initial Velocity

In this scenario, the stone is dropped rather than being thrown, so the initial velocity ( \(v_0y\) ) will be 0 m/s.
04

Apply given values in the equation

Let's insert these values in the given equation: \((10 m/s)^2 = (0 m/s)^2 - 2 * (9.8 m/s^2) * (10m - 0m)\)
05

Solve the Equation

Solving this equation the right side now becomes (-196), which can't be true because the square of velocity can never be negative. So there must be a mistake in the equation or it could represent getting at a certain speed (10 m/s) before hitting the ground, starting from a certain height (10m) and achieving this without the effect of gravity.

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 91影视!

Key Concepts

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

Free Fall
When objects move under the influence of gravity alone, without any other force acting on them, they are said to be in "free fall."
Free fall assumes no air resistance, so the only force is gravitational pull.
It doesn鈥檛 matter what weights are, they all accelerate at the same rate due to gravity in free fall.
  • This acceleration is always present when free-falling objects are considered.
  • There's no initial velocity pushing the object sideways or upwards.
Think of free fall like when you drop a ball from your hand. The only thing making it move is gravity pulling it towards the ground.
It鈥檚 vital to understand that because of gravity, the velocity of the object will increase the longer it falls.
Gravity
Gravity is the natural force that makes objects pulled towards the Earth.
It鈥檚 a crucial concept in understanding how objects move.
For free-falling objects, gravity provides a constant acceleration:
  • It鈥檚 denoted by the symbol \( g \) and typically, it's quantified as \( 9.8 \, m/s^2 \).
  • This means that in each second of free fall, the velocity of an object increases by \( 9.8 \, m/s \).
Gravity is why planes stay up, or why you don鈥檛 float off into space.
In physics problems, it鈥檚 typically seen as acting downwards, towards the center of the Earth.
Velocity Calculation
Velocity describes how fast an object is moving in a specific direction.
In free fall, we often calculate the final velocity before an object hits the ground.
Here's how you deal with velocity calculation:
  • When calculating velocity, it's crucial to set up known values like initial velocity, which might be \( 0 \, m/s \) if dropped.
  • Solving the velocity often involves using kinematic equations to relate velocity, acceleration, and displacement.
Remember, initial velocity is the speed before free fall begins, and the final velocity is what you often calculate.
In these situations, always double-check units to make sure they鈥檙e consistent throughout the problem.
Kinematic Equations
Kinematic equations are tools used to describe the motion of objects.
They are super helpful to predict how objects move under certain conditions.
Here's a breakdown:
  • They relate various aspects of motion like displacement, initial and final velocity, time, and acceleration.
  • In the equation from the problem, [(10 m/s)虏 = (v_0 y)虏 - 2(9.8 m/s虏)(10 m - 0 m)], you are setting up the relationships between these variables.
These equations are used to not only solve the final velocity but also to understand the full range of an object's motion.
Knowing how to apply these correctly gives you the power to analyze motion realistically and accurately.

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

You want to visit your friend in Seattle during spring break. To save money, you decide to travel there by train. Unfortunately, your physics final exam took the full 3 hours, so you are Late in arriving at the train station. You run as fast as you can, but just as you reach the platform you see your train, \(30 \mathrm{m}\) ahead of you down the platform, begin to accelerate at \(1.0 \mathrm{m} / \mathrm{s}^{2} .\) You chase after the train at your maximum speed of \(8.0 \mathrm{m} / \mathrm{s},\) but there's a barrier \(50 \mathrm{m}\) ahead. Will you be able to leap onto the back step of the train before you crash into the barrier?

A car starts from rest at a stop sign. It accelerates at \(4.0 \mathrm{m} / \mathrm{s}^{2}\) for \(6.0 \mathrm{s},\) coasts for \(2.0 \mathrm{s},\) and then slows down at a rate of \(3.0 \mathrm{m} / \mathrm{s}^{2}\) for the next stop sign. How far apart are the stop signs?

A driver has a reaction time of \(0.50 \mathrm{s}\), and the maximum deceleration of her car is \(6.0 \mathrm{m} / \mathrm{s}^{2} .\) She is driving at \(20 \mathrm{m} / \mathrm{s}\) when suddenly she sees an obstacle in the road \(50 \mathrm{m}\) in front of her. Can she stop the car in time to avoid a collision?

You are given the kinematic equation or equations that are used to solve a problem. For each of these, you are to: a. Write a realistic problem for which this is the correct equation(s). Be sure that the answer your problem requests is consistent with the equation(s) given. b. Draw the pictorial representation for your problem. c. Finish the solution of the problem. $$64 \mathrm{m}=0 \mathrm{m}+(32 \mathrm{m} / \mathrm{s})(4 \mathrm{s}-0 \mathrm{s})+\frac{1}{2} a_{x}(4 \mathrm{s}-0 \mathrm{s})^{2}$$

You're driving down the highway late one night at \(20 \mathrm{m} / \mathrm{s}\) when a deer steps onto the road \(35 \mathrm{m}\) in front of you. Your reaction time before stepping on the brakes is \(0.50 \mathrm{s}\), and the maximum deceleration of your car is \(10 \mathrm{m} / \mathrm{s}^{2}\) a. How much distance is between you and the deer when you come to a stop? b. What is the maximum speed you could have and still not hit the deer?
See all solutions

Recommended explanations on Physics Textbooks

Turning Points in Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Space Physics

Read Explanation

Wave Optics

Read Explanation

Electricity and Magnetism

Read Explanation

Dynamics

Read Explanation
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.