/*! 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 48 A large boulder is ejected verti... [FREE SOLUTION] | 91Ó°ÊÓ

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A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 \(\mathrm{m} / \mathrm{s} .\) Air resistance may be ignored. (a) At what time after being ejected is the boulder moving at 20.0 \(\mathrm{m} / \mathrm{s}\) upward? (b) At what time is it moving at 20.0 \(\mathrm{m} / \mathrm{s}\) downward? (c) When is the displacement of the boulder from its initial position zero? (d) When is the velocity of the boulder zero? (e) What are the magnitude and direction of the acceleration while the boulder is (i) moving upward? (ii) Moving downward? (iii) At the highest point? (f) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-\) t graphs for the motion.

Short Answer

Expert verified
(a) 2.04 s, (b) 6.12 s, (c) 4.08 s, (d) 4.08 s, (e) -9.8 m/s² for all. Graphs reflect these relations.

Step by step solution

01

Identify the Given and Required Values

We know that the initial velocity \( u = 40.0 \, \text{m/s} \). We need to find: (a) the time \( t_1 \) when the velocity is 20.0 m/s upward, (b) the time \( t_2 \) when the velocity is 20.0 m/s downward, (c) the time \( t_3 \) when the displacement is zero, (d) the time \( t_4 \) when the velocity is zero, (e) the magnitude and direction of the acceleration at various stages, (f) sketches of the related graphs.
02

Equation of Motion

For an object under uniform acceleration and ignoring air resistance, we use the equation \( v = u + at \), where \( v \) is the final velocity, \( u \) is the initial velocity, \( a = -9.8 \, \text{m/s}^2 \) due to gravity, and \( t \) is the time.
03

Calculate Time for 20.0 m/s Upward

Using the equation \( v = u + at \), solve for \( t_1 \): \[ 20 = 40 - 9.8t_1 \] Rearranging gives: \[ t_1 = \frac{40 - 20}{9.8} = 2.04 \, \text{s} \]
04

Calculate Time for 20.0 m/s Downward

When the boulder reaches 20.0 m/s downward, its velocity is negative. Use the equation again: \[ -20 = 40 - 9.8t_2 \] Rearranging gives: \[ t_2 = \frac{40 + 20}{9.8} = 6.12 \, \text{s} \]
05

Calculate Time When Displacement is Zero

The time when displacement is zero corresponds to the total time of flight. Since it starts and ends at the same position, use the symmetry of projectile motion: \[ t_3 = 2 \times t_4 = 4.08 \, \text{s} \] Here, \( t_4 \) from part (d) is the time to reach the maximum height (zero velocity).
06

Calculate Time When Velocity is Zero

The velocity is zero at the highest point. Solve \( 0 = 40 - 9.8t_4 \): \[ t_4 = \frac{40}{9.8} = 4.08 \, \text{s} \]
07

Analyze Acceleration During Motion

Regardless of the direction of motion, acceleration due to gravity is constant at \( -9.8 \, \text{m/s}^2 \). This includes: (i) Moving upward: \( a = -9.8 \, \text{m/s}^2 \), (ii) Moving downward: \( a = -9.8 \, \text{m/s}^2 \), (iii) At the highest point: \( a = -9.8 \, \text{m/s}^2 \).
08

Sketch the Motion Graphs

While not visually sketchable here, 1. The \( a_y-t \) graph is a horizontal line at \( -9.8 \, \text{m/s}^2 \). 2. The \( v_y-t \) graph is a straight line descending from 40 m/s to -40 m/s. 3. The \( y-t \) graph is parabolic upwards, peaking at the maximum height.

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Key Concepts

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

Kinematics Equations
In physics, kinematics equations are vital to studying the motion of objects. These equations help us understand how objects move under the influence of forces like gravity. For projectile motion, like a boulder ejected from a volcano, the kinematic equation we'll often use is:\[ v = u + at \]Where:
  • \(v\) is the final velocity.
  • \(u\) is the initial velocity.
  • \(a\) is the acceleration due to gravity, which is \(-9.8 \, \text{m/s}^2\) for objects near the Earth's surface.
  • \(t\) is the time elapsed.
This equation allows us to calculate how fast an object is moving at any point during its ascent or descent, assuming we're ignoring air resistance. It is a powerful tool in analyzing projectile motion because it links the key variables we often need to solve for in these types of problems.
Acceleration due to Gravity
Gravity is a force that pulls objects toward the center of the Earth. In freefall motion, this force acts as a constant acceleration. This acceleration is always the same near the Earth's surface, at \(-9.8 \, \text{m/s}^2\). It's crucial to remember that the negative sign here indicates that gravity opposes the upward motion of a projectile.When a boulder is thrown upwards from a volcano, its upward velocity decreases over time due to this gravitational pull. Eventually, the boulder reaches a point where its velocity becomes zero at the peak of its travel, before it starts descending back to the ground, accelerating under gravity.Hence, whether the boulder is moving up, going down, or even at the peak, gravity remains a consistent, unchanging force. This constancy makes calculations more straightforward, allowing us to predict how objects move in a gravitational field with confidence.
Freefall Motion
When an object is in freefall, it is only under the influence of gravity with no other forces, such as air resistance, acting on it. This condition simplifies the analysis of its motion. Initially, when launched upwards, the boulder's velocity decreases by \(9.8 \, \text{m/s}\) every second until it reaches its peak. Then, as it falls back down, it gains velocity at the same rate.The boulder reaches its maximum height when its velocity becomes zero. At this moment, although the boulder is momentarily not moving upward or downward, gravity does not stop acting on it. The entire journey from launch to peak height and back to its original level is a perfect example of symmetrically balanced freefall motion.Understanding freefall is key in predicting how long the boulder takes to rise and fall, and determining total time in the air, using the total flight symmetry and other factors.
Velocity-Time Graphs
Velocity-time graphs are a powerful visual tool for understanding motion. When plotting the velocity of the boulder against time, we get a straight line, showing a decrease from its initial velocity to zero, and continuing to a negative velocity as it falls back to the ground.- Initially, the velocity is high as the boulder is launched upwards.- The slope of the line represents the constant acceleration due to gravity, at \(-9.8 \, \text{m/s}^2\), showing a steady decline to zero.- As the boulder descends, the graph continues linearly, dipping into negative values.This linear graph tells us that the velocity changes uniformly with time, confirming the constant nature of gravitational acceleration. It's not only a method of visualization but also helps in calculating critical points, like when the boulder's velocity is zero or when it changes direction, all of which are essential for fully grasping projectile motion dynamics.

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Most popular questions from this chapter

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CALC A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by \(x(t)=b t^{2}-c t^{3},\) where \(b=2.40 \mathrm{m} / \mathrm{s}^{2}\) and \(c=0.120 \mathrm{m} / \mathrm{s}^{3} .\) (a) Calculate the average velocity of the car for the time interval \(t=0\) to \(t=10.0\) s. (b) Calculate the instantaneous velocity of the car at \(t=0, t=5.0 \mathrm{s},\) and \(t=10.0 \mathrm{s}\) . (c) How long after starting from rest is the car again at rest?

A turtle crawls along a straight line, which we will call the \(x\) -axis with the positive direction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{cm}+\) \((2.00 \mathrm{cm} / \mathrm{s}) t-\left(0.0625 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2}\) (a) Find the turtle's initial velocity, initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of 10.0 \(\mathrm{cm}\) from its starting point? What is the velocity (magnitude and direction) of the turtle at each of these times? (e) Sketch graphs of \(x\) versus \(t, v_{x}\) versus \(t,\) and \(a_{x}\) versus \(t\) for the time interval \(t=0\) to \(t=40 \mathrm{s}.\)

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