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Chapter 2: Problem 48

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) 8.16 s; (d) 4.08 s; (e) \(-9.8\, \mathrm{m/s^2}\) always.

Step by step solution

01

Identify the Known Values and Equations

The initial velocity of the boulder is given as \( v_0 = 40.0 \, \mathrm{m/s} \). Since the boulder is in free fall, the acceleration due to gravity is \( a = -9.8 \, \mathrm{m/s^2} \). Use the equation \( v = v_0 + at \) to find the time \( t \) when the velocity \( v \) equals the given values in parts a and b.
02

Find Time for Part (a) - Boulder Moving Upward at 20 m/s

For part (a), solve the equation \( 20.0 = 40.0 - 9.8t \) to find the time. Simplifying gives:\[ 20.0 = 40.0 - 9.8t \]\[ 9.8t = 20.0 \]\[ t = \frac{20.0}{9.8} \approx 2.04 \, \mathrm{s} \].
03

Find Time for Part (b) - Boulder Moving Downward at 20 m/s

For part (b), when the boulder is moving downward, its velocity is -20.0 m/s. Solve the equation \( -20.0 = 40.0 - 9.8t \):\[ 40.0 - 9.8t = -20.0 \]\[ 9.8t = 60.0 \]\[ t = \frac{60.0}{9.8} \approx 6.12 \, \mathrm{s} \].
04

Determine When Displacement is Zero

Displacement is zero when the boulder reaches the same vertical level from which it was ejected. The time for a full trajectory (up and down back to the starting point) can be calculated by finding when \( v = 0 \) first, determining the time to the highest point, and then doubling it. The time to the highest point is when the velocity \( v = 0 \) using \( 0 = 40.0 - 9.8t \):\[ 9.8t = 40.0 \]\[ t = \frac{40.0}{9.8} \approx 4.08 \, \mathrm{s} \].So, the full round trip time is \( 2 \times 4.08 = 8.16 \, \mathrm{s} \).
05

Determine When Velocity is Zero

The velocity is zero at the highest point of the trajectory. Use the result from Step 4: \( t = 4.08 \, \mathrm{s} \).
06

Determine Acceleration in All Situations

The acceleration due to gravity \( a = -9.8 \, \mathrm{m/s^2} \) is constant:- (i) While moving upward: \( a = -9.8 \, \mathrm{m/s^2} \).- (ii) While moving downward: \( a = -9.8 \, \mathrm{m/s^2} \).- (iii) At the highest point: \( a = -9.8 \, \mathrm{m/s^2} \).
07

Sketch Graphs

Three graphs represent the motion:1. \( a_y-t \): A horizontal line at \( a = -9.8 \, \mathrm{m/s^2} \).2. \( v_y-t \): A line with a negative slope, starting at 40 m/s and crossing zero at 4.08 s heading to -40 m/s at 8.16 s.3. \( y-t \): A parabola opening downward, starting from the origin, peaking at 4.08 s and returning to zero at 8.16 s.

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Key Concepts

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

Projectile Motion
Projectile motion is a form of kinematics where an object is launched into the air and influenced solely by the force of gravity. In the case of the boulder from the original exercise, it is launched vertically upward, which simplifies the projectile motion to a one-dimensional problem.
When dealing with projectile motion, it's important to note several key characteristics:
  • Initial velocity: The boulder begins with an initial velocity of 40.0 m/s. This is the speed at which it is ejected from the volcano.
  • Path of motion: While it moves upward, gravity acts to decelerate it until it reaches its peak height, then gravity accelerates it downward.
  • Symmetrical motion: The time it takes to ascend to the highest point equals the time to descend back to the starting point, assuming no air resistance.
  • Velocity and position: Velocity decreases as it ascends, momentarily becomes zero at the peak, and then increases in the downward direction.
Understanding these characteristics helps in solving problems related to vertical projection, such as calculating the time it takes for the boulder to reach specific velocities during its journey.
Free Fall
Free fall describes the motion of an object where gravity is the only force acting upon it. In the case of the boulder, because air resistance is ignored, it is in free fall.
Common attributes of free fall include:
  • Acceleration: The acceleration due to gravity, denoted as \( g \), is constant at approximately \( 9.8 \, \mathrm{m/s^2} \) downward.
  • Independence of mass: All objects, regardless of their mass, experience the same acceleration if air resistance is negligible.
  • Velocity changes: As the boulder rises, its velocity decreases until it stops. As it falls back, its velocity increases.
Free fall allows the use of simple kinematic equations to predict the motion's parameters, such as time, velocity, and displacement. For example, determining the time at which the boulder鈥檚 upward velocity is 20.0 m/s or when it reaches a downward velocity of -20.0 m/s.
Acceleration due to Gravity
Acceleration due to gravity is a key concept in understanding motion, especially in vertical scenarios like our boulder problem. This acceleration is considered a constant value, \( 9.8 \, \mathrm{m/s^2} \), and acts downward towards the center of the Earth.
In the vertical projection example, this acceleration:
  • Affects motion equally: Whether the boulder is moving up, at rest at the peak, or moving down, the acceleration due to gravity remains the same at \( -9.8 \, \mathrm{m/s^2} \).
  • Influences velocity: It reduces upward velocity until it is zero at the peak and increases the downward velocity thereafter.
  • Consistency: No matter where on its path the boulder is, gravity's influence is constant and unidirectional.
This concept simplifies calculations and predictions as we can assume the same acceleration value at all points in the motion, making equations easier to solve and graphs such as the velocity-time graph linear.
Velocity-Time Graph
A velocity-time graph for vertical projectile motion, like that of the volcano-ejected boulder, provides a visual representation of how velocity changes over time.
Key aspects of the velocity-time graph include:
  • Slope interpretation: The slope of the graph is the acceleration. In this case, it is constant at \( -9.8 \, \mathrm{m/s^2} \), signified by a straight line with a downward slope.
  • Changes in velocity: The graph starts at \( 40.0 \, \mathrm{m/s} \), drops to zero at the highest point around 4.08 seconds, and continues downward to \( -40.0 \, \mathrm{m/s} \) by around 8.16 seconds.
  • Area under the graph: This area gives the displacement. A positive area indicates upward motion, while a negative area represents downward motion.
This graph becomes a practical tool for visualizing moments of zero velocity and maximum displacement, helping to intuitively understand events during the boulder's flight, such as how long it takes to reach the peak and return.

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Most popular questions from this chapter

Two cars, \(A\) and \(B\) , travel in a straight line. The distance of \(A\) from the starting point is given as a function of time by \(x_{A}(t)=\alpha t+\beta t^{2},\) with \(\alpha=2.60 \mathrm{m} / \mathrm{s}\) and \(\beta=1.20 \mathrm{m} / \mathrm{s}^{2} .\) The distance of \(B\) from the starting point is \(x_{B}(t)=\gamma t^{2}-\delta t^{3},\) with \(\gamma=2.80 \mathrm{m} / \mathrm{s}^{2}\) and \(\delta=0.20 \mathrm{m} / \mathrm{s}^{3} .\) (a) Which car is ahead just after they leave the starting point? (b) At what time(s) are the cars at the same point? (c) At what time(s) is the distance from \(A\) to \(B\) neither increasing nor decreasing? (d) At what time(s) do \(A\) and \(B\) have the same acceleration?

On a 20 mile bike ride, you ride the first 10 miles at an average speed of 8 \(\mathrm{mi} / \mathrm{h}\) . What must your average speed over the next 10 miles be to have your average speed for the total 20 miles be (a) 4 \(\mathrm{mi} / \mathrm{h} ?\) (b) 12 \(\mathrm{mi} / \mathrm{h} ?(\mathrm{c})\) Given this average speed for the first 10 miles, can you possibly attain an average speed of 16 \(\mathrm{mi} / \mathrm{h}\) for the total 20 -mile ride? Explain.

A turtle crawls along a straight line, which we will call the \(x\) -axis with the positive dircction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{cm}+\) \((2.00 \mathrm{cm} / \mathrm{s}) t-\left(0.0625 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2},(\mathrm{a})\) Find the turte'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 \(t=40 \mathrm{s}\)

Relay Race. In a relay race, each contestant runs 25.0 \(\mathrm{m}\) while carrying an egg balanced on a spoon, turns around, and comes back to the starting point. Edith runs the first 25.0 \(\mathrm{m}\) in 20.0 \(\mathrm{s}\) . On the return trip she is more confident and takes only 15.0 \(\mathrm{s}\) . What is the magnitude of her average velocity for \((\mathrm{a})\) the first 25.0 \(\mathrm{m} ?\) (b) The return trip? (c) What is her average velocity for the entire round trip? (d) What is her average speed for the round trip?

Building Height. Spider-Man steps from the top of a tall building. He falls freely from rest to the ground a distance of \(h .\) He falls a distance of \(h / 4\) in the last 1.0 \(\mathrm{s}\) of his fall. What is the height \(h\) of the building?
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