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A rocket carrying a satellite is accelerating straight up from the earth's surface. At 1.15 s after liftoff, the rocket clears the top of its launch platform, 63 \(\mathrm{m}\) above the ground. After an additional \(4.75 \mathrm{s},\) it is 1.00 \(\mathrm{km}\) above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75 -s part of its flight and (b) the first 5.90 s of its flight.

Short Answer

Expert verified
(a) 197.47 m/s, (b) 169.49 m/s.

Step by step solution

01

Understand the Problem

We need to find the average velocity of a rocket in two parts of its flight. First, calculate the average velocity for the time interval between 1.15 s after liftoff to an additional 4.75 s (totaling 5.90 s). Second, calculate for the entire 5.90 s after liftoff.
02

Write Down Known Values

The rocket is 63 m above the ground 1.15 s after liftoff and 1000 m (1 km) above the ground after an additional 4.75 s. The total time for the second part is 5.90 s after liftoff.
03

Calculate Average Velocity for Part 1

For the first part, distance covered is from 63 m to 1000 m, in 4.75 seconds. Use the formula for average velocity: \[ v_{avg} = \frac{\Delta s}{\Delta t} = \frac{s_{final} - s_{initial}}{t_{final} - t_{initial}} \] Substituting the known values gives: \[ v_{avg} = \frac{1000\, \text{m} - 63\, \text{m}}{4.75\, \text{s}} \] \[ v_{avg} = \frac{937\, \text{m}}{4.75\, \text{s}} \approx 197.47\, \text{m/s} \]
04

Calculate Average Velocity for Part 2

For the entire 5.90 seconds, distance from 0 m to 1000 m in 5.90 seconds. Using the average velocity formula again: \[ v_{avg} = \frac{1000\, \text{m} - 0\, \text{m}}{5.90\, \text{s}} \] \[ v_{avg} = \frac{1000\, \text{m}}{5.90\, \text{s}} \approx 169.49\, \text{m/s} \]

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Key Concepts

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

Rocket Acceleration
Rockets use powerful engines to blast themselves into space. This involves the concept of **acceleration**, which is how quickly an object's speed changes over time. For a rocket, this acceleration needs to be quite intense to overcome Earth's gravity. Every second that passes, the rocket gets faster, making the acceleration crucial to escape gravitational pull.

Rockets start their journey from rest, which means their initial velocity is zero. As time progresses, the engines provide a thrust, causing the rocket's velocity to increase continuously. This change in velocity over time defines the acceleration of the rocket.

If we consider the physics of rocket motion, broad factors such as force, mass, and acceleration all interlink here. According to **Newton's Second Law of Motion**:
  • Force equals mass times acceleration (\(F = ma\)).
  • This can be rearranged to find acceleration: \(a = \frac{F}{m}\).
This means the heavier the rocket, the more thrust (force) it needs to accelerate.

Real-life rocket launches take into account the mass of the rocket, the thrust provided by the engines, and how this changes with time as fuel burns off, reducing the total mass of the vehicle. Understanding these principles can lead to more efficient cosmic journeys.
Displacement Calculation
Displacement refers to the change in position of an object. For the rocket, displacement is measured in how far it moves vertically from its starting point on the launchpad.
To determine displacement, consider:
  • Initial Position: Where was the rocket when you started timing?
  • Final Position: Where is the rocket after moving for some time?
  • Time Interval: How much time has passed during this motion?
In the context of the original exercise, displacement was calculated by measuring the rocket's position at different times.

For instance, at 1.15 seconds after liftoff, the rocket has moved 63 meters up. After another 4.75 seconds, it reaches a height of 1000 meters. The actual path may be complex, but for displacement, only the start and end points matter, not the actual trajectory traveled.
The formula used here is:\[\text{Displacement} = s_{final} - s_{initial}\]This signifies simply subtracting the starting position from the final position. It helps in understanding the movement concerning the original point of departure.
Velocity Formula
Velocity is an important aspect of understanding motion, especially for something as dynamic as a rocket. It shows not just how fast an object is moving (speed), but also in what direction.

To get the average velocity, it's necessary to calculate how much ground is covered over a period of time. The formula for **average velocity** is:\[v_{avg} = \frac{\Delta s}{\Delta t} = \frac{s_{final} - s_{initial}}{t_{final} - t_{initial}}\]Where:
  • \(\Delta s\) is the change in position, or displacement.
  • \(\Delta t\) is the change in time.
In the exercise, the rocket's average velocity was calculated for two different time intervals: the 4.75 seconds from after clearing the launch platform and the entire 5.90 seconds from launch.

By using the formula, you can see which part of the journey the rocket was traveling faster. The average speeds of 197.47 m/s and 169.49 m/s, respectively, provide insights into how quickly the rocket's mission progresses. Understanding velocity gives valuable insights into the efficiency and effectiveness of the rocket's launch and propulsion phases.

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Most popular questions from this chapter

A subway train starts from rest at a station and accelerates at a rate of 1.60 \(\mathrm{m} / \mathrm{s}^{2}\) for 14.0 \(\mathrm{s}\) . It runs at constant speed for 70.0 \(\mathrm{s}\) and slows down at a rate of 3.50 \(\mathrm{m} / \mathrm{s}^{2}\) until it stops at the next station. Find the total distance covered.

On a 20 -mile bike ride, you ride the first 10 miles at an average speed of 8 \(\mathrm{mi} / \mathrm{h} .\) What must your average speed over the next 10 miles be to have your average speed for the total 20 miles be (a) 4 \(\mathrm{mi} / \mathrm{h} ?\) (b) 12 \(\mathrm{mi} / \mathrm{h} ?\) (c) Given this average speed for the first 10 miles, can you possibly attain an average speed of 16 \(\mathrm{mi} / \mathrm{h}\) for the total 20 -mile ride? Explain.

Catching the Bus. A student is running at her top speed of 5.0 \(\mathrm{m} / \mathrm{s}\) to catch a bus, which is stopped at the bus stop. When the student is still 40.0 \(\mathrm{m}\) from the bus, it starts to pull away, moving with a constant acceleration of 0.170 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) For how much time and what distance does the student have to run at 5.0 \(\mathrm{m} / \mathrm{s}\) before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling? (c) Sketch an \(x\) -t graph for both the student and the bus. Take \(x=0\) at the initial position of the student. (d) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus traveling at this point? (e) If the student's top speed is 3.5 \(\mathrm{m} / \mathrm{s},\) will she catch the bus? (f) What is the minimum speed the student must have to just catch up with the bus? For what time and what distance does she have to run in that case?

Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. (a) If the faster stone takes 10 s to return to the ground, how long will it take the slower stone to return? (b) If the slower stone reaches a maximum height of \(H\) , how high (in terms of \(H )\) will the faster stone go? Assume free fall.

Launch of the Space Shuttle. At launch the space shuttle weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 \(\mathrm{km} / \mathrm{h}\) , and at the end of the first 1.00 \(\mathrm{min}\) its speed is 1610 \(\mathrm{km} / \mathrm{h}\) . (a) What is the average acceleration (\(\operatorname{in}\) \(\mathrm{m} / \mathrm{s}^{2}\) ) of the shuttle (i) during the first \(8.00 \mathrm{s},\) and (ii) between 8.00 \(\mathrm{s}\) and the end of the first 1.00 \(\mathrm{min}\) ? (b) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals, what distance does the shuttle travel (i) during the first \(8.00 \mathrm{s},\) and ( ii) during the interval from 8.00 s to 1.00 \(\mathrm{min}\)?

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