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A spacecraft is traveling with a velocity of \(v_{0 x}=5480 \mathrm{m} / \mathrm{s}\) along the \(+x\) direction. Two engines are turned on for a time of 842 s. One engine gives the spacecraft an acceleration in the \(+x\) direction of \(a_{x}=1.20 \mathrm{m} / \mathrm{s}^{2}\), while the other gives it an acceleration in the \(+y\) direction of \(a_{y}=8.40 \mathrm{m} / \mathrm{s}^{2}\) At the end of the firing, find (a) \(v_{x}\) and (b) \(v_{y^{*}}\)

Short Answer

Expert verified
(a) \(v_x = 6490.4 \ \text{m/s}\), (b) \(v_y = 7072.8 \ \text{m/s}\)

Step by step solution

01

Understanding the Given Information

We have a spacecraft initially traveling with a velocity of \(v_{0 x} = 5480 \ \text{m/s}\) in the \(+x\) direction. The spacecraft experiences an additional acceleration in the \(+x\) direction, \(a_x = 1.20 \ \text{m/s}^2\), and an acceleration in the \(+y\) direction, \(a_y = 8.40 \ \text{m/s}^2\). The accelerations last for \(t = 842 \ \text{s}\). We need to find the final velocities in the \(+x\) and \(+y\) directions after this time duration.
02

Calculate the Final Velocity in the x-direction

The final velocity in the \(x\) direction, \(v_x\), can be calculated using the kinematic equation: \(v_x = v_{0x} + a_x \cdot t\). Substitute \(v_{0x} = 5480 \ \text{m/s}\), \(a_x = 1.20 \ \text{m/s}^2\), and \(t = 842 \ \text{s}\).
03

Plug in the Values for v_x

Substitute the values into the equation: \(v_x = 5480 + 1.20 \times 842\). Calculate \(1.20 \times 842 = 1010.4\). Then, add this value to the initial velocity: \(v_x = 5480 + 1010.4 = 6490.4 \ \text{m/s}\).
04

Calculate the Final Velocity in the y-direction

The final velocity in the \(y\) direction, \(v_y\), is calculated similarly using the equation: \(v_y = v_{0y} + a_y \cdot t\). Here, \(v_{0y} = 0 \ \text{m/s}\) since the spacecraft initially had no velocity in the \(y\) direction.
05

Plug in the Values for v_y

Using the values \(v_{0y} = 0\), \(a_y = 8.40 \ \text{m/s}^2\), and \(t = 842 \ \text{s}\), substitute them into the equation: \(v_y = 0 + 8.40 \times 842\). Calculate \(8.40 \times 842 = 7072.8\). Therefore, \(v_y = 7072.8 \ \text{m/s}\).

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Key Concepts

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

Velocity
When we talk about velocity, we're referring to how fast something moves in a specific direction. Imagine a spacecraft zipping through space; it not only travels at some speed but also heads toward a particular direction. This is exactly what velocity encapsulates—it includes both the speed and direction of motion.
Velocity is a vector quantity, which means it has both magnitude (how fast the object is going) and direction (where it's headed). In the exercise, the spacecraft initially moves with a velocity of 5480 m/s in the +x direction. As it undergoes acceleration due to the engines, this initial velocity will change over time. Let's detail this below:
  • **Magnitude**: The speed part of velocity. For the spacecraft, the initial speed is 5480 m/s in the +x direction.
  • **Direction**: Here, the spacecraft starts in the +x direction. Any movement in the y-direction will also be considered when determining its total velocity.
Velocity changes due to acceleration. As the engines provide additional acceleration in both the x and y directions, the spacecraft's velocity in these directions will change, resulting in new final velocities after a given period.
Acceleration
Acceleration tells us how quickly the velocity of an object is changing. It's another vector quantity, which means it has both magnitude and direction. In simpler terms, acceleration is how fast something speeds up, slows down, or changes direction. In our space scenario, the spacecraft is influenced by two types of acceleration:
  • **In the +x direction**: The acceleration here is 1.20 m/s², meaning that every second, the velocity of the spacecraft in the x direction increases by 1.20 m/s.
  • **In the +y direction**: The spacecraft also experiences a lateral acceleration of 8.40 m/s², which means its velocity in the y direction increases by 8.40 m/s each second.
These accelerations combine over the 842 seconds the engines are active, significantly altering both the speed and direction of the spacecraft's motion. Acceleration ensures that the velocity of an object is not static and continuously varies under these parameters.
Kinematic Equations
Kinematic equations are the mathematical tools that help us describe the motion of objects. They relate the five primary motion variables: initial velocity, final velocity, acceleration, time, and displacement. These equations are tremendously useful in calculating the final velocities of our spacecraft.
For our case, we used the equation for final velocity:
  • In the x-direction: \(v_x = v_{0x} + a_x imes t\)
  • In the y-direction: \(v_y = v_{0y} + a_y imes t\)
Here’s what each term means:
  • **\(v_{0x}\) and \(v_{0y}\)**: Initial velocities in the x and y directions, respectively.
  • **\(a_x\) and \(a_y\)**: Acceleration in the x and y directions, respectively.
  • **\(t\)**: The total time the accelerations are applied.
In our solution, these formulas allow us to calculate the new final velocities for the spacecraft, taking into account both the time the engines are on and the provided accelerations. These equations simplify the complex motion into manageable calculations, making it easier to understand how all the factors interact in kinematics.

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Most popular questions from this chapter

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