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Chapter 4: Problem 120

A rocket lifts off the surface of Earth with a constant acceleration of \(20 \mathrm{m} / \mathrm{sec}^{2} .\) How fast will the rocket be going 1 min later?

Short Answer

Expert verified
The rocket will be going 1200 m/s.

Step by step solution

01

Understand the Problem

We need to find the final velocity of the rocket after 1 minute of liftoff, given a constant acceleration of \(20 \, \mathrm{m/s^2}\).
02

Identify Known Values

- Initial velocity (\(v_0\)) is \(0 \, \mathrm{m/s}\) since the rocket is at rest at the start.- Acceleration (\(a\)) is \(20 \, \mathrm{m/s^2}\).- Time (\(t\)) is \(1\) minute, which we need to convert to seconds: \(1 \, \text{minute} = 60 \, \text{seconds}\).
03

Choose the Correct Formula

We use the formula for final velocity with constant acceleration: \[ v = v_0 + a imes t \]Where:- \(v\) is the final velocity,- \(v_0\) is the initial velocity,- \(a\) is the acceleration,- \(t\) is the time.
04

Substitute the Known Values

Substitute \(v_0 = 0 \, \mathrm{m/s}\), \(a = 20 \, \mathrm{m/s^2}\), and \(t = 60 \, \mathrm{s}\) into the formula:\[ v = 0 + 20 imes 60 \]
05

Calculate the Final Velocity

Calculate:\[ v = 20 imes 60 = 1200 \, \mathrm{m/s} \]
06

Conclusion

The rocket will be traveling at a speed of \(1200 \, \mathrm{m/s}\) after 1 minute of liftoff.

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Key Concepts

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

Kinematics
Kinematics is the branch of mechanics that studies the motion of objects without considering the forces that cause the motion. It's primarily focused on parameters like displacement, velocity, acceleration, and time. Kinematics gives us the tools to describe motion through equations and graphs.

The main variables often considered in kinematics are:
  • Displacement (\( s \)): The change in position of an object.
  • Velocity (\( v \)): The rate of change of displacement with time.
  • Acceleration (\( a \)): The rate of change of velocity with time.
  • Time (\( t \)): The duration over which motion occurs.
Understanding kinematics is crucial for analyzing and predicting the motion of objects, such as the trajectory of a rocket launched into space. When dealing with constant acceleration, such as in the exercise you encountered, a direct relationship allows us to calculate final velocity if the initial velocity and time are known.
Final Velocity Formula
In kinematics, the final velocity of an object is determined by how its velocity changes over time due to acceleration. With constant acceleration, you can calculate the final velocity using the formula:\[ v = v_0 + a \times t\]

This formula shows that final velocity (\(v\)) depends on:
  • Initial velocity (\(v_0\)): The starting speed of the object. For instance, if the object starts from rest, \(v_0\) is zero.
  • Acceleration (\(a\)): The constant rate at which the object's velocity changes over time. It's given in \(\mathrm{m/s^2}\).
  • Time (\(t\)): The duration of time over which the acceleration is applied, usually converted to seconds if given in minutes.
Using the final velocity formula allows us to find how fast an object will move after a certain period, assuming a constant rate of acceleration. In our exercise, it was applied to calculate how fast the rocket goes after 1 minute of constant acceleration.
Time Conversion
Time conversion plays an essential role in solving problems involving equations of motion, as the units of time need to be consistent with those of other quantities like acceleration. Often, problems present time in different units, and it's crucial to convert these units for proper application in formulas.

For the exercise, the given time was in minutes, but kinematic equations typically require time in seconds. Here's a simple way to convert:
  • 1 minute is equivalent to 60 seconds.
Converting units helps ensure accuracy in calculations. When you input time in seconds for the formula, it matches the \(\mathrm{m/s^2}\) units of acceleration, allowing for a seamless operation and correct final result. Always pay attention to units when performing calculations; conversion is a frequent but straightforward task in solving physics problems.

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Most popular questions from this chapter

a. The U.S. Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 108 in. What dimensions will give a box with a square end the largest possible volume? GRAPH CANT COPY b. Graph the volume of a 108 -in. box (length plus girth equals 108 in.) as a function of its length and compare what you see with your answer in part (a). (Continuation of Exercise \(20 .\) ) a. Suppose that instead of having a box with square ends you have a box with square sides so that its dimensions are \(h\) by \(h\) by \(w\) and the girth is \(2 h+2 w .\) What dimensions will give the box its largest volume now?b. Graph the volume as a function of \(h\) and compare what you see with your answer in part (a). GRAPH CANT COPY

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Sketch a smooth connected curve \(y=f(x)\) with \(\begin{aligned}f(-2) &=8 \\\f(0) &=4 \\\f(2) &=0 \\\f^{\prime}(x) &>0 \quad \text { for } \quad|x|>2\end{aligned}\) \(\begin{aligned}&f^{\prime}(2)=f^{\prime}(-2)=0\\\&f^{\prime}(x)<0 \text { for }|x|<2\\\&f^{\prime \prime}(x)<0 \text { for } x<0\\\&f^{\prime \prime}(x)>0 \quad \text { for } \quad x>0\end{aligned}\)
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