/*! 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 25 On a spacecraft, two engines are... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

On a spacecraft, two engines are turned on for 684 \({s}\) at a moment when the velocity of the craft has \(x\) and \(y\) components of \(v_{0 x}=4370 {m} / {m} / {s}\) and \(v_{0 y}=6280 {m} / {s}\) s. While the engines are firing, the craft undergoes a displacement that has components of \(x=4.11 \times 10^{6} {m}\) and \(y=6.07 \times 10^{6} {m} .\) Find the \(x\) and \(y\) components of the craft's acceleration.

Short Answer

Expert verified
The spacecraft's acceleration components are \(a_x = 4.78 \, \text{m/s}^2\) and \(a_y = 7.59 \, \text{m/s}^2\).

Step by step solution

01

Identify Given Values

We begin by identifying the initial conditions and known variables from the problem. We have:- Initial velocity components: \(v_{0x} = 4370 \, \text{m/s} \) \(v_{0y} = 6280 \, \text{m/s} \)- Displacement components: \(x = 4.11 \times 10^{6} \, \text{m} \) \(y = 6.07 \times 10^{6} \, \text{m} \)- Time duration of engine firing: \( t = 684 \, \text{s} \)
02

Use Kinematic Equation for Acceleration

The kinematic equation relating displacement, initial velocity, time, and acceleration is:\[ s = v_0 t + \frac{1}{2} a t^2 \]This needs to be applied separately to the x and y components.
03

Solve for x-component of Acceleration

For the x-component, we use:\[ x = v_{0x} t + \frac{1}{2} a_x t^2 \]Plug in the known values:\[ 4.11 \times 10^6 = 4370 \times 684 + \frac{1}{2} a_x \times 684^2 \]Simplify and solve for \(a_x\):\[ 4.11 \times 10^6 = 2.99028 \times 10^6 + 234156 a_x \]\[ 1.11972 \times 10^6 = 234156 a_x \]\[ a_x = \frac{1.11972 \times 10^6}{234156} \approx 4.78 \, \text{m/s}^2 \]
04

Solve for y-component of Acceleration

For the y-component, we use:\[ y = v_{0y} t + \frac{1}{2} a_y t^2 \]Plug in the known values:\[ 6.07 \times 10^6 = 6280 \times 684 + \frac{1}{2} a_y \times 684^2 \]Simplify and solve for \(a_y\):\[ 6.07 \times 10^6 = 4.29352 \times 10^6 + 234156 a_y \]\[ 1.77648 \times 10^6 = 234156 a_y \]\[ a_y = \frac{1.77648 \times 10^6}{234156} = 7.59 \, \text{m/s}^2 \]
05

Conclusion: Final Acceleration Components

The x and y components of the spacecraft's acceleration are:- \(a_x = 4.78 \, \text{m/s}^2\)- \(a_y = 7.59 \, \text{m/s}^2\)

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.

Acceleration
Acceleration is a fundamental concept in kinematics that describes the rate of change of velocity over time. In simpler terms, it's how quickly an object speeds up or slows down. For the spacecraft in the problem, the engines firing cause a change in velocity, thereby leading to acceleration. In this exercise, we find that the spacecraft has two acceleration components: one along the x-axis and one along the y-axis. These components tell us how the speed in each direction changes due to the engines. For example, a higher acceleration value in the x direction means the spacecraft is speeding up more quickly horizontally.
  • The formula for finding acceleration components is derived from the kinematic equation.
  • The accelerations are calculated by comparing initial velocity, displacement, and time.
Velocity Components
Velocity components refer to the parts of an object's velocity that occur in specific directions. In the context of kinematics, it's often helpful to break down velocity into horizontal (x-component) and vertical (y-component) parts.The initial velocity components given in the exercise are:
  • Horizontal velocity, \( v_{0x} = 4370 \, \text{m/s}\)
  • Vertical velocity, \( v_{0y} = 6280 \, \text{m/s}\)
These values indicate how fast the spacecraft was moving horizontally and vertically before the engines were fired. By considering the separate components, it becomes easier to manage and solve for how each direction changes independently during the motion. This separation into components is crucial when analyzing movements in different dimensions.
Displacement
Displacement is the vector quantity that describes the change in position of an object. It contains both the magnitude and the direction of the movement.In the given problem, the spacecraft's displacement components are provided as:
  • Horizontal displacement, \( x = 4.11 \times 10^6 \, \text{m} \)
  • Vertical displacement, \( y = 6.07 \times 10^6 \, \text{m} \)
These values show how far the spacecraft has moved along the x and y axes while the engines were engaged. The displacement helps to calculate the effect of the firing engines over time, showing how much ground is covered in both the horizontal and vertical axes. Understanding displacement is key to analyzing motion, as it directly relates to finding out how velocity changes.
Kinematic Equations
Kinematic equations are mathematical formulas used to describe the motion of objects under constant acceleration. They relate displacement, velocity, acceleration, and time, making them invaluable for solving problems like the one in the exercise.The specific equation used here is:\[ s = v_0 t + \frac{1}{2} a t^2 \]This equation helps calculate an object's displacement based on its initial velocity, acceleration, and the time period over which these occur.
  • The equation is applied separately to each direction: x and y.
  • It enables solving for unknown acceleration by rearranging the formula.
By leveraging the kinematic equations, one can determine acceleration values when certain components like initial velocity, displacement, and time are known. This approach simplifies the understanding and calculation of complex motions.

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

SSM A skateboarder, starting from rest, rolls down a 12.0-m ramp. When she arrives at the bottom of the ramp her speed is 7.70 m/s. (a) Determine the magnitude of her acceleration, assumed to be constant. (b) If the ramp is inclined at 25.0 with respect to the ground, what is the component of her acceleration that is parallel to the ground?

MMH A golfer, standing on a fairway, hits a shot to a green that is elevated 5.50 m above the point where she is standing. If the ball leaves her club with a velocity of 46.0 m/s at an angle of 35.0 above the ground, find the time that the ball is in the air before it hits the green.

At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt on which you can either stand or walk. Suppose a speed ramp has a length of 105 m and is moving at a speed of 2.0 m/s relative to the ground. In addition, suppose you can cover this distance in 75 s when walking on the ground. If you walk at the same rate with respect to the speed ramp that you walk on the ground, how long does it take for you to travel the 105 m using the speed ramp?

In the absence of air resistance, a projectile is launched from and returns to ground level. It follows a trajectory similar to that shown in Figure 3.10 and has a range of 23 m. Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?

Stones are thrown horizontally with the same velocity from the tops of two different buildings. One stone lands twice as far from the base of the building from which it was thrown as does the other stone. Find the ratio of the height of the taller building to the height of the shorter building.

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.