/*! 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 51 A rocket starts from rest and mo... [FREE SOLUTION] | 91Ó°ÊÓ

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A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by \(a_{y}=\left(2.80 \mathrm{m} / \mathrm{s}^{3}\right) t,\) where acceleration of the rocket is given by \(a_{y}=\left(2.80 \mathrm{m} / \mathrm{s}^{3}\right) t,\) where the \(+y\) -direction is upward. (a) What is the height of the rocket above the surface of the earth at \(t=10.0 \mathrm{s} ?\) (b) What is the speed of the rocket when it is 325 \(\mathrm{m}\) above the surface of the earth?

Short Answer

Expert verified
At \( t = 10 \ \text{s} \), the height is 466.67 m. The speed at 325 m is 110.13 m/s.

Step by step solution

01

Identify Known Variables and Equation

The rocket's upward acceleration is given by \( a_y = (2.80 \ \text{m/s}^3)t \), and the initial conditions are \( v_0 = 0 \ \text{m/s} \) and \( y_0 = 0 \ \text{m} \). We need to solve for the height at \( t = 10 \ \text{s} \) and the speed when the rocket is 325 m above the surface.
02

Find Velocity Function

To find the velocity function, we need to integrate the acceleration function with respect to time. The acceleration function is \( a(t) = (2.80 \ \text{m/s}^3)t \). So, \( v(t) = \int a(t) \ dt = \int (2.80 \ t) \ dt = 1.40 \ t^2 + C \), where \( C \) is the integration constant. Since \( v(0) = 0 \), it follows that \( C = 0 \). Thus, \( v(t) = 1.40 \ t^2 \).
03

Find Position Function

To find the position, integrate the velocity function with respect to time. \( y(t) = \int v(t) \ dt = \int (1.40 \ t^2) \ dt = \frac{1.40}{3}t^3 + C \). With \( y(0) = 0 \), \( C = 0 \). Hence, \( y(t) = \frac{1.40}{3}t^3 \).
04

Calculate Height at \(t = 10\ \text{s}\)

Substitute \( t = 10 \) into the position function to find the height: \( y(10) = \frac{1.40}{3}(10)^3 = \frac{1.40}{3}(1000) = 466.67 \ \text{m} \).
05

Solve for Speed at 325 m Above Surface

From the position function \( y = \frac{1.40}{3}t^3 \), solve for \( t \) when \( y = 325 \text{ m}\): \( 325 = \frac{1.40}{3} t^3 \Rightarrow t^3 = \frac{325 \times 3}{1.40} = 696.43 \Rightarrow t = \sqrt[3]{696.43} \approx 8.86 \ \text{s} \).
06

Calculate Speed at \(t = 8.86\ \text{s}\)

Substitute \( t \approx 8.86 \) into the velocity function \( v(t) = 1.40 \ t^2 \): \( v(8.86) = 1.40 \ (8.86)^2 \approx 110.13 \ \text{m/s} \).

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Key Concepts

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

Rocket Motion
Rocket motion is a classic example in kinematics that helps illustrate various principles of physics. It involves studying how a rocket travels through space under the influence of forces such as gravity and thrust. In this exercise, we look at a rocket that starts from rest on the surface of the Earth and moves upward, demonstrating accelerated motion.
The motion of the rocket is driven by its engines, which provide thrust and generate vertical acceleration. During the rocket’s ascent, the only significant force acting on it is the thrust from the engines, assuming air resistance is negligible at this stage. This exercises focuses on understanding how the rocket gains height and speed with time, using given acceleration data.
Analyzing the motion involves using initial conditions such as starting speed and position, which are both zero in this case. This allows us to calculate how these parameters change, giving us insight into the rocket's behavior over time.
Vertical Acceleration
Vertical acceleration refers to the rate of change of velocity in the upward direction. In this example, the rocket has an acceleration given by the function \(a_y = (2.80 \ ext{m/s}^3) t\).
This means that the acceleration is not constant, but rather it increases linearly with time. Specifically, as time progresses, the acceleration becomes stronger, signifying that the engines are continuously providing more thrust.
The given function implies a model where acceleration depends on time, indicating a possibly simplified physics problem suited for educational purposes. In real life, various factors such as fuel consumption, changing mass, and atmospheric drag would also play a role.
In kinematics problems dealing with non-constant acceleration like this one, important steps involve integrating the acceleration function to determine the velocity and position over time. This process unfolds in subsequent sections, where we calculate these values at specific times.
Position Function
The position function describes the displacement of the rocket from its initial position after a certain time. The position of the rocket as a function of time is obtained by integrating the velocity function.
The integration of the velocity function \(v(t) = 1.40 \ t^2\) leads to the position function: \[y(t) = \frac{1.40}{3} t^3\].
This position equation allows us to calculate how high the rocket has traveled above the Earth's surface over time. For example, at \(t = 10\) seconds, the calculated height is approximately 466.67 meters.
  • Knowing the position function helps in various practical applications, such as determining whether the rocket is following its intended trajectory or calculating the time to reach certain altitudes.
  • It further provides a visualization of the relationship between acceleration, velocity, and displacement, showing how these quantities evolve over time during the motion.
The position function is crucial for predicting future positions based on current and past data, enabling effective planning and navigation in rocketry.
Velocity Function
The velocity function tells us how the speed of an object changes with time during its motion. In the problem, we find the velocity function of the rocket by integrating the given acceleration function.
Starting with the acceleration \(a(t) = (2.80 \ ext{m/s}^3)t\), integrate it to determine the velocity: \[v(t) = 1.40 \ t^2\].
This function shows that the velocity is proportional to the square of the time, indicating that the speed of the rocket increases quadratically. The function provides an expression that can calculate the rocket’s speed at any given moment.
The velocity function is helpful in understanding how fast the rocket is moving at any point during its ascent. For instance, when the rocket reaches 325 meters in height, solving for time and using it in the velocity function gives a speed of approximately 110.13 m/s.
  • Such calculations are pivotal when ensuring that a rocket doesn't exceed safe speed limits or when assessing the effectiveness of propulsion systems.
  • It also aids in adjustments needed mid-flight to maintain the desired course or velocity.
Through the velocity function, we gain insights into the dynamics of rocket propulsion and motion.

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Most popular questions from this chapter

A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes the pot 0.420 s to pass from the top to the bottom of this window, which is 1.90 \(\mathrm{m}\) high. How far is the top of the window below the windowsill from which the flowerpot fell?

Determined to test the law of gravity for himself, a student walks off a skyscraper 180 \(\mathrm{m}\) high, stopwatch in hand, and starts his free fall (zero initial velocity). Five seconds later, Superman arrives at the scene and dives off the roof to save the student. Superman leaves the roof with an initial speed \(v_{0}\) that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. (a) What must the value of \(v_{0}\) be so that Superman catches the student just before they reach the ground? (b) On the same graph, sketch the positions of the student and of Superman as functions of time. Take Superman's initial speed to have the value calculated in part (a). (c) If the height of the skyscraper is less than some minimum value, even Superman can't reach the student before he hits the ground. What is this minimum height?

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