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Chapter 2: Problem 58

(II) A rocket rises vertically, from rest, with an acceleration of \(3.2 \mathrm{~m} / \mathrm{s}^{2}\) until it runs out of fuel at an altitude of \(950 \mathrm{~m}\) After this point, its acceleration is that of gravity, downward. \((a)\) What is the velocity of the rocket when it runs out of fuel? \((b)\) How long does it take to reach this point? (c) What maximum altitude does the rocket reach? \((d)\) How much time (total) does it take to reach maximum altitude? (e) With what velocity does it strike the Earth? \((f)\) How long (total) is it in the air?

Short Answer

Expert verified
(a) 78 m/s, (b) 24.375 s, (c) 1260 m, (d) 32.335 s, (e) 157 m/s, (f) 48.365 s.

Step by step solution

01

Identify the Given Information

The rocket has an initial velocity of \(u = 0 \text{ m/s}\), an upward acceleration \(a = 3.2 \text{ m/s}^2\), and it runs out of fuel at an altitude \(s = 950 \text{ m}\). After this point, the acceleration is gravitational, \(g = 9.8 \text{ m/s}^2\). This information will help us solve all parts of the problem.
02

Calculate Velocity When Fuel Runs Out (a)

Using the kinematic equation \(v^2 = u^2 + 2as\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the distance, calculate the velocity:\[v^2 = 0 + 2 \times 3.2 \times 950\]\[v^2 = 6080\]\[v = \sqrt{6080} \approx 78 \text{ m/s}\]Therefore, the velocity when it runs out of fuel is approximately \(78 \text{ m/s}\).
03

Calculate Time to Fuel Exhaustion (b)

Using the equation \(v = u + at\), solve for time \(t\):\[78 = 0 + 3.2t\]\[t = \frac{78}{3.2} \approx 24.375 \text{ seconds}\]Thus, it takes approximately \(24.375\) seconds to reach this point.
04

Calculate Maximum Altitude Reached (c)

When the rocket runs out of fuel, it continues upwards against the gravitational force until its velocity becomes zero. Using the equation \(v^2 = u^2 + 2as\) with upward motion, where acceleration \(a = -9.8 \text{ m/s}^2\), and initial velocity from Step 2:\[0 = 78^2 + 2(-9.8)s\]\[s = \frac{78^2}{2 \times 9.8} \approx 310 \text{ m}\]Adding this to the altitude at fuel exhaustion: \(950 + 310 = 1260 \text{ m}\).The maximum altitude reached by the rocket is \(1260 \text{ m}\).
05

Time to Reach Maximum Altitude (d)

Calculate the additional time taken from fuel exhaustion to maximum altitude using \(v = u + at\):\[0 = 78 + (-9.8)t\]\[t = \frac{78}{9.8} \approx 7.96 \text{ seconds}\]Total time to reach maximum altitude: \(24.375 + 7.96 = 32.335 \text{ seconds}\).
06

Calculate Velocity Upon Earth Impact (e)

To find the velocity with which the rocket strikes the Earth when it falls from 1260 meters, use \(v^2 = u^2 + 2as\) where \(u=0\), \(a=9.8 \text{ m/s}^2\), and \(s=1260\):\[v^2 = 0 + 2 \times 9.8 \times 1260\]\[v^2 = 24696\]\[v = \sqrt{24696} \approx 157 \text{ m/s}\].Thus, the velocity as it strikes the Earth is approximately \(157 \text{ m/s}\).
07

Total Time in the Air (f)

Calculate the time taken to fall from the maximum altitude back to Earth using \(s = ut + \frac{1}{2}at^2\), where initial velocity \(u = 0\), gravitational acceleration \(a = 9.8 \text{ m/s}^2\), and \(s = 1260\):\[1260 = \frac{1}{2} \times 9.8 \times t^2\]\[t^2 \approx 257.14\]\[t \approx \sqrt{257.14} \approx 16.03 \text{ seconds}\].Total time in the air: \(32.335 + 16.03 = 48.365 \text{ seconds}\).

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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 an exciting concept that involves understanding the journey of a rocket from launch to landing. When a rocket launches, it may start from rest and move upwards due to the force generated by burning fuel. This upward motion is characterized by positive acceleration, which propels the rocket higher and higher.
Once the fuel is exhausted, the rocket's motion is influenced solely by gravity, which means it experiences downward acceleration. This shift dramatically impacts how high the rocket can go and how it moves after reaching its peak altitude.
Summarizing rocket motion involves familiar topics:
  • Initial acceleration due to propulsion.
  • Transition to gravity-driven motion once fuel ends.
  • Effects of different forces on the rocket's trajectory.
Understanding these principles gives insights into how we send rockets into space and safely return them.
Acceleration Due to Gravity
Acceleration due to gravity is a fundamental concept in physics, particularly in kinematics. On Earth, this acceleration is typically constant at approximately \(9.8 \, \text{m/s}^2\). This constant acceleration acts downwards, towards the center of the Earth, and affects all objects in freefall.
For the rocket, after its fuel is depleted, this gravitational force becomes the only acceleration acting on it. This results in the upward movement slowing down until it stops and eventually changes direction towards the Earth.
The implications of gravitational acceleration include:
  • It's always active, influencing objects regardless of their motion.
  • Freefall conditions when no other forces like air resistance act on the object.
  • A critical factor in calculating time of flight and velocity of descending objects.
Hence, understanding gravity is crucial, not just for the journey of any rocket, but for all motion involving objects on Earth.
Kinematic Equations
Kinematic equations provide essential tools for solving problems related to motion, including those involving rockets. In the case of constant acceleration, these equations allow us to connect key motion variables like velocity, acceleration, time, and displacement.
For example, the equation \(v^2 = u^2 + 2as\) helps determine the final velocity of a rocket when it runs out of fuel. This equation relates initial velocity \(u\), acceleration \(a\), and displacement \(s\) to the final velocity \(v\).
Some core kinematic equations include:
  • \(v = u + at\): Calculates velocity change over time.
  • \(s = ut + \frac{1}{2}at^2\): Determines displacement over time.
  • \(v^2 = u^2 + 2as\): Links velocity to displacement under constant acceleration.
By mastering these equations, students can solve diverse problems in physics, enhancing their understanding of motion in one dimension.

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Most popular questions from this chapter

(II) In coming to a stop, a car leaves skid marks \(85 \mathrm{~m}\) long on the highway. Assuming a deceleration of \(4.00 \mathrm{~m} / \mathrm{s}^{2},\) estimate the speed of the car just before braking.

(I) A car accelerates from \(12 \mathrm{~m} / \mathrm{s}\) to \(21 \mathrm{~m} / \mathrm{s}\) in \(6.0 \mathrm{~s}\). What was its acceleration? How far did it travel in this time? Assume constant acceleration.

(II) A car moving in a straight line starts at \(x=0\) at \(t=0\). It passes the point \(x=25.0 \mathrm{~m}\) with a speed of \(11.0 \mathrm{~m} / \mathrm{s}\) at \(t=3.00 \mathrm{~s} .\) It passes the point \(x=385 \mathrm{~m}\) with a speed of \(45.0 \mathrm{~m} / \mathrm{s}\) at \(t=20.0 \mathrm{~s}\). Find \((a)\) the average velocity and (b) the average acceleration between \(t=3.00 \mathrm{~s}\) and \(t=20.0 \mathrm{~s}\)

(II) The Table below gives the speed of a particular drag racer as a function of time. (a) Calculate the average acceleration \(\left(\mathrm{m} / \mathrm{s}^{2}\right)\) during each time interval. \((b)\) Using numerical integration (see Section \(2-9\) ) estimate the total distance traveled \((\mathrm{m})\) as a function of time. [Hint: for \(\bar{v}\) in each interval sum the velocities at the beginning and end of the interval and divide by \(2 ;\) for example, in the second interval use \(\bar{v}=(6.0+13.2) / 2=9.6]\) (c) Graph each of these. $$ \begin{array}{lccccccccccc} \hline t(\mathrm{~s}) & 0 & 0.50 & 1.00 & 1.50 & 2.00 & 2.50 & 3.00 & 3.50 & 4.00 & 4.50 & 5.00 \\ v(\mathrm{~km} / \mathrm{h}) & 0.0 & 6.0 & 13.2 & 22.3 & 32.2 & 43.0 & 53.5 & 62.6 & 70.6 & 78.4 & 85.1 \\ \hline \end{array} $$

You are traveling at a constant speed \(v_{\mathrm{M}},\) and there is a car in front of you traveling with a speed \(v_{\mathrm{A}}\). You notice that \(v_{M}>v_{\mathrm{A}},\) so you start slowing down with a constant acceleration \(a\) when the distance between you and the other car is \(x\). What relationship between \(a\) and \(x\) determines whether or not you run into the car in front of you?
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