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Chapter 2: Problem 32

Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initial velocity of \(+5800 \mathrm{m} / \mathrm{s},\) while rocket \(\mathrm{B}\) has an initial velocity of \(+8600 \mathrm{m} / \mathrm{s}\). After a time \(t\) both rockets are again side by side, the displacement of each being zero. The acceleration of rocket \(A\) is \(-15 \mathrm{m} / \mathrm{s}^{2}\). What is the acceleration of rocket B?

Short Answer

Expert verified
The acceleration of rocket B is \(-21.6 \, \mathrm{m/s^2}\).

Step by step solution

01

Identify the Known Variables

We know the initial velocity of rocket A is \( v_{0A} = 5800 \, \mathrm{m/s} \), the initial velocity of rocket B is \( v_{0B} = 8600 \, \mathrm{m/s} \), and the acceleration of rocket A is \( a_A = -15 \, \mathrm{m/s^2} \). The displacement for both rockets when they are again side by side is zero.
02

Understand the Condition

Both rockets have zero displacement when they are again side by side. This means that each rocket returns to its starting position after some time \( t \). We can use the formula for displacement which is \( s = v_0 t + \frac{1}{2} a t^2 \) to set up equations for both rockets.
03

Write Equation for Rocket A

The displacement equation for rocket A is \( 0 = 5800t + \frac{1}{2}(-15)t^2 \). Simplifying, \( 0 = 5800t - 7.5t^2 \).
04

Solve for Time using Rocket A

From the equation \( 5800t - 7.5t^2 = 0 \), factor out \( t \): \( t(5800 - 7.5t) = 0 \). \( t = 0 \) is when they are side by side initially, so we need to solve \( 5800 = 7.5t \). Solving gives \( t = \frac{5800}{7.5} \approx 773.33 \, \mathrm{s}\).
05

Write Equation for Rocket B

The displacement equation for rocket B is \( 0 = 8600t + \frac{1}{2}a_B t^2 \). We have found \( t \approx 773.33 \, \mathrm{s} \), so substituting, \( 0 = 8600(773.33) + \frac{1}{2} a_B (773.33)^2 \).
06

Solve for the Acceleration of Rocket B

Substitute the value of \( t \) in the equation \( 0 = 8600 \times 773.33 + \frac{1}{2} a_B \times (773.33)^2 \). This becomes \(0 = 8600 \times 773.33 + 0.5 \times a_B \times 597635.89 \). Solving, \( a_B = -21.6 \, \mathrm{m/s^2}\).

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Key Concepts

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

Rocket Motion
Rocket motion involves the study of how rockets travel through space. This complex motion is governed by principles of physics, helping us to predict their paths. When rockets move, they often have engines that expel gas at high speeds. This causes thrust, pushing the rocket forward. In this exercise, two rockets are initially moving side by side. Understanding their motion is crucial for predicting their future positions. Key Points about Rocket Motion:
  • The initial velocity is the speed at which the rocket starts moving. For Rocket A, it was 5800 m/s, and for Rocket B, 8600 m/s.
  • Acceleration becomes important when a rocket changes how fast it's moving. In this problem, when their retrorockets fire, the rockets experience acceleration.
  • The principle of displacement, which we'll explore, shows us how these rockets return where they started after a period.
Acceleration Calculation
Acceleration describes how quickly an object changes its velocity. In rocket science, it's critical because small changes in speed can substantially affect a rocket's path.To calculate acceleration, we often need to set up equations based on known and unknown values, like initial velocity and time.

Steps to Calculate Acceleration

  • Write the displacement formula: \( s = v_0 t + \frac{1}{2} a t^2 \). This formula helps determine how much a rocket moves over time.
  • Substitute known values to solve for unknowns. In our exercise, Rocket A's given values were used to find the time it took for both rockets to be side by side again.
  • Use the same formula for Rocket B, substituting the time found from Rocket A's equation to solve for Rocket B's acceleration, resulting in \( a_B = -21.6 \, \mathrm{m/s^2} \).
Displacement in Physics
Displacement in physics refers to the change in position of an object. It tells us how far out of place an object is.For rockets, understanding displacement is essential because it shows us their path in space.

Understanding Displacement

  • When displacement is zero, as in our problem, it means the object returns to its original position. This happens after the retrorockets of both rockets fire.
  • The formula for displacement \( s = v_0 t + \frac{1}{2} a t^2 \) combines initial velocity, time, and acceleration to calculate how far an object moves.
  • In our exercise, using displacement equations for each rocket, we could find when they were side by side again and calculate important variables like acceleration.
By breaking down these concepts, students can better understand the principles behind rocket motion and the calculations involved.

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Most popular questions from this chapter

In getting ready to slam-dunk the ball, a basketball player starts from rest and sprints to a speed of \(6.0 \mathrm{m} / \mathrm{s}\) in \(1.5 \mathrm{s}\). Assuming that the player accelerates uniformly, determine the distance he runs.

A hot-air balloon is rising upward with a constant speed of \(2.50 \mathrm{m} / \mathrm{s}\). When the balloon is \(3.00 \mathrm{m}\) above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?

A top-fuel dragster starts from rest and has a constant acceleration of \(40.0 \mathrm{m} / \mathrm{s}^{2}\). Concepts: (i) At time \(t\) the dragster has a certain velocity. Keeping in mind that the dragster starts from rest, when the time doubles to \(2 t\), does the velocity also double? (ii) When the time doubles to \(2 t,\) does the displacement of the dragster also double? Calculations: What are the (a) final velocities and (b) displacements of the dragster at the end of \(2.0 \mathrm{s}\) and at the end of twice this time, or \(4.0 \mathrm{s} ?\)

Over a time interval of 2.16 years, the velocity of a planet orbiting a distant star reverses direction, changing from \(+20.9 \mathrm{km} / \mathrm{s}\) to \(-18.5 \mathrm{km} / \mathrm{s} .\) Find (a) the total change in the planet's velocity (in \(\mathrm{m} / \mathrm{s}\) ) and (b) its average acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.

From her bedroom window a girl drops a water-filled balloon to the ground, \(6.0 \mathrm{m}\) below. If the balloon is released from rest, how long is it in the air?
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