/*! 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 44 When startled, an armadillo will... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When startled, an armadillo will leap upward. Suppose it rises \(0.544 \mathrm{~m}\) in the first \(0.200 \mathrm{~s}\). (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of \(0.544 \mathrm{~m}\) ? (c) How much higher does it go?

Short Answer

Expert verified
(a) Initial speed is 3.701 m/s. (b) Speed at 0.544 m is 1.74 m/s. (c) It goes 0.698 m higher.

Step by step solution

01

Identify Given Information

We are given that the armadillo rises to a height of \(0.544 \ \mathrm{m}\) in \(0.200 \ \mathrm{s}\). We also know the acceleration due to gravity, \(g = 9.81 \ \mathrm{m/s^2}\), acts downward.
02

Find Initial Speed (Part a)

To find the initial speed \(v_0\), use the kinematic equation for distance: \[ s = v_0 t + \frac{1}{2} a t^2 \]Here, \(s = 0.544 \ \mathrm{m}\), \(t = 0.200 \ \mathrm{s}\), and \(a = -9.81 \ \mathrm{m/s^2}\). Solve for \(v_0\):\[ 0.544 = v_0 (0.200) + \frac{1}{2} (-9.81) (0.200)^2 \]\[ 0.544 = 0.200v_0 - 0.1962 \]\[ 0.7402 = 0.200v_0 \]\[ v_0 = \frac{0.7402}{0.200} = 3.701 \ \mathrm{m/s} \]
03

Calculate Speed at Maximum Height (Part b)

At maximum height, the final velocity \(v\) is 0, but for the height 0.544 m use:\[ v^2 = v_0^2 + 2a s \]At \(s = 0.544 \ \mathrm{m}\), solve for \(v\):\[ v^2 = 3.701^2 + 2(-9.81)(0.544) \]\[ v^2 = 13.696 + (-10.66944) \]\[ v^2 = 3.02656 \]\[ v = \sqrt{3.02656} = 1.74 \ \mathrm{m/s} \]
04

Determine Additional Height (Part c)

At \(0 \ \mathrm{m/s}\), write:\[ 0 = 3.701^2 + 2(-9.81) h_\text{additional} \]\[ 0 = 13.696 + (-19.62) h_\text{additional} \]\[ 19.62 h_\text{additional} = 13.696 \]\[ h_\text{additional} = \frac{13.696}{19.62} = 0.698 \ \mathrm{m} \]Thus, the armadillo goes higher by \(0.698 \ \mathrm{m}\).

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.

Initial Velocity
When talking about the initial velocity in a kinematic problem, we're referring to the speed at which an object starts its motion. For our armadillo problem, initial velocity (\(v_0\)) is the speed right as the armadillo leaves the ground. Knowing this is crucial because it allows us to predict the motion path of the armadillo thereafter.
To determine this initial velocity, we use the kinematic equation for motion with constant acceleration:\[ s = v_0 t + \frac{1}{2} a t^2 \]where \(s\) is the distance traveled, \(t\) is the time, and \(a\) is the acceleration due to gravity. Remember, gravity acts downwards and is always negative in our equations.

In our exercise, the armadillo rises to 0.544 meters in just 0.200 seconds. Plugging these values into our equation lets us solve for \(v_0\). It's a powerful technique to determine beginning speed just by knowing how far something goes in a certain time span.
Kinematic Equations
Kinematic equations are the backbone of analyzing motion in physics. They describe the movement of objects in various conditions like time, acceleration, and velocity. The cornerstone equations include:
  • \[ s = v_0 t + \frac{1}{2} a t^2 \]helps calculate the distance moved.
  • \[ v = v_0 + at \]which gives final velocity based on time and acceleration.
  • \[ v^2 = v_0^2 + 2as \]useful for finding velocity when time isn't directly involved.
In the context of our armadillo exercise, we use these equations to:- Determine the initial speed.- Find how fast it's going at a certain height.- Calculate additional height it achieves.

Our versatile kinematic equations let us dig into parts of motion that aren’t immediately visible, like intermediate speeds or heights reached. Recognizing when and how to use these formulas is key—not only for solving textbook problems but in practical situations like sports or engineering scenarios too.
Gravitational Acceleration
Gravitational acceleration (\(g\)) is a constant value which dictates how fast an object accelerates toward Earth. On Earth, \(g\) is approximately \(9.81 \, \text{m/s}^2\) pointing downwards. This constant force impacts everything in free fall, like our armadillo.

Understanding \(g\) is crucial, as the direction (usually negative in upward motion scenarios) and size of this acceleration affects how objects behave when they leap, fall, or are thrown.
In our example with the armadillo, \(g\) diminishes the initial velocity as the animal travels upward until it’s momentarily stationary at the peak of its jump. Then, gravity pulls it back down, increasing its speed until it hits the ground again.
  • Gravitational force is a constant influencer in any kind of vertical motion.
  • Because it's always vertical, the biggest trick is getting the direction right in your calculations.
Grasping this concept of gravity and its impact is essential for not only solving kinematic equations but also understanding how things move in our daily life.

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

A ball of moist clay falls \(15.0 \mathrm{~m}\) to the ground. It is in contact with the ground for \(20.0 \mathrm{~ms}\) before stopping. (a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? (Treat the ball as a particle.) (b) Is the average acceleration up or down?

The Zero Gravity Research Facility at the NASA Glenn Research Center includes a \(145 \mathrm{~m}\) drop tower. This is an evacuated vertical tower through which, among other possibilities, a 1 -m-diameter sphere containing an experimental package can be dropped. (a) How long is the sphere in free fall? (b) What is its speed just as it reaches a catching device at the bottom of the tower? (c) When caught, the sphere experiences an average deceleration of \(25 g\) as its speed is reduced to zero. Through what distance does it travel during the deceleration?

A shuffleboard disk is accelerated at a constant rate from rest to a speed of \(6.0 \mathrm{~m} / \mathrm{s}\) over a \(1.8 \mathrm{~m}\) distance by a player using a cue. At this point the disk loses contact with the cue and slows at a constant rate of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) until it stops. (a) How much time elapses from when the disk begins to accelerate until it stops? (b) What total distance does the disk travel?

Water drips from the nozzle of a shower onto the floor 200 \(\mathrm{cm}\) below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. When the first drop strikes the floor, how far below the nozzle are the (a) second and (b) third drops?

In 1889 , at Jubbulpore, India, a tug-of-war was finally won after 2 h 41 min, with the winning team displacing the center of the rope \(3.7 \mathrm{~m}\). In centimeters per minute, what was the magnitude of the average velocity of that center point during the contest?

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.