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A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of \(384,000 \mathrm{km}\) . Suppose the spaceship starts from rest and accelerates at 20.0 \(\mathrm{m} / \mathrm{s}^{2}\) for the first 15.0 min of the trip, and then travels at constant speed until the last 15.0 min, when it slows down at a rate of 20.0 \(\mathrm{m} / \mathrm{s}^{2}\) , just coming to rest as it reaches the moon. (a) What is the maximum speed attained? (b) What fraction of the total distance is traveled at constant speed? (c) What total time is required for the trip?

Short Answer

Expert verified
(a) 18,000 m/s (b) 95.8% (c) 6.18 hours

Step by step solution

01

Calculate Time of Acceleration and Deceleration in Seconds

The spaceship accelerates and decelerates for 15 minutes each. Since there are 60 seconds in a minute, we convert 15 minutes to seconds: \[15 ext{ minutes} = 15 imes 60 = 900 ext{ seconds}.\]
02

Calculate Maximum Speed Attained

Using the formula for final velocity under constant acceleration, \( v = u + at \), where the initial velocity \( u = 0 \), acceleration \( a = 20 \, \text{m/s}^2 \), and time \( t = 900 \, \text{seconds} \), we calculate:\[v = 0 + (20)(900) = 18,000 \, \text{m/s}.\]Thus, the maximum speed attained is \( 18,000 \, \text{m/s} \).
03

Calculate Distance Covered During Acceleration and Deceleration

The distance covered during acceleration can be computed with the formula for distance under constant acceleration: \( s = ut + \frac{1}{2}at^2 \). For uniform acceleration:\[s = 0 + \frac{1}{2}(20)(900)^2 = 8,100,000 \, \text{m} = 8,100 \, \text{km}. \]The same calculation applies for the distance during deceleration since it's symmetric. Therefore, each phase covers \( 8,100 \, \text{km} \).
04

Calculate Remaining Distance Traveled at Constant Speed

The total distance to the moon is \( 384,000 \, \text{km} \). The total distance covered during acceleration and deceleration phases is \( 16,200 \, \text{km} \). Thus, the distance covered at constant speed is:\[384,000 - 16,200 = 367,800 \, \text{km}.\]
05

Calculate Time Taken to Travel at Constant Speed

We know the speed during this phase is the maximum speed attained, \( 18,000 \, \text{m/s} \). Convert distance into meters: \( 367,800 \, \text{km} = 367,800,000 \, \text{m} \).The time \( t \) taken is given by the formula: \( t = \frac{d}{v} \), where \( d \) is the distance, and \( v \) is velocity:\[t = \frac{367,800,000}{18,000} = 20,433.33 \, \text{seconds}.\]
06

Calculate Total Travel Time

The total travel time is the sum of the acceleration time, cruising time, and deceleration time:\[Total \ time = 900 \, \text{seconds (acceleration)} + 20,433.33 \, \text{seconds (cruising)} + 900 \, \text{seconds (deceleration)} = 22,233.33 \, \text{seconds}\]Convert to hours: \( \frac{22,233.33}{3600} \approx 6.18 \text{ hours}. \)
07

Calculate Fraction of Total Distance at Constant Speed

The fraction of the total distance traveled at constant speed is given by the distance traveled at constant speed divided by the total distance:\[\text{Fraction} = \frac{367,800}{384,000} = \frac{612}{640} \approx 0.958.\]

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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 physics that deals with the study of motion, without considering the forces that cause the motion. It focuses on understanding how objects move, looking at parameters like displacement, velocity, acceleration, and time. When solving kinematic problems, we often use specific equations, known as kinematic equations, to describe the motion of objects. These equations relate the four fundamental motion variables: initial velocity, final velocity, acceleration, time, and displacement.

In the case of the spaceship problem, we use these equations to determine how the spaceship moves from Earth to the Moon. In scenarios involving straight-line motion, like this, the equations become simplified. Each phase of the trip (acceleration, constant speed, and deceleration), can be tackled using the relevant kinematic equations. Understanding kinematics is crucial for predicting an object's future position and speed based on its current state and the forces applied.
Constant Acceleration
Constant acceleration refers to a situation in which an object's acceleration does not change over time. This means that the velocity of the object increases uniformly in linear motion. Constant acceleration is an important concept in kinematics because it creates a predictable pattern in the change of velocity and allows us to use specific equations to easily calculate other variables.

In our problem, the spaceship experiences constant acceleration at the start and end of its journey, accelerating at a rate of 20 m/s². This is crucial because it enables us to use the equation: \[ v = u + at \]where \( v \) is final velocity, \( u \) is initial velocity, \( a \) is acceleration, and \( t \) is time. Since the spaceship starts from rest (\( u = 0 \)), we can derive its maximum speed (\( v \)) after a given time period. This approach simplifies complex motion problems and is widely used in physics.
Motion in One Dimension
Motion in one dimension refers to the movement of objects along a straight line. This type of motion is the simplest form, as it involves changes in position or speed along a single axis. Most one-dimensional motion is addressed using the kinematic equations, as they help us understand how time, velocity, displacement, and acceleration interrelate in linear paths.

For the spaceship, its journey to the moon is an example of one-dimensional motion. Despite the vast distances and high speeds involved, the problem reduces to simple calculations in a single direction—there are no turns or changes in path. This single directional focus makes it easier to apply kinematic equations, resulting in predictable and precise calculations of different motion attributes.
Uniform Motion
Uniform motion occurs when an object travels at a constant speed, covering equal distances in equal amounts of time. In this state, there is no change in velocity, meaning that acceleration is zero. Uniform motion is a special case of motion that can simplify calculations significantly, particularly when evaluating systems over longer periods of sustained speed.

The spaceship achieves uniform motion after its initial acceleration phase when it reaches its maximum speed. During this phase, it travels most of the distance between the Earth and the Moon at this constant speed. This allows us to use a simple formula for distance: \[ s = vt \]where \( s \) is distance, \( v \) is velocity, and \( t \) is time. This simplicity aids in understanding how long the spaceship travels at maximum velocity and what fraction of the total trip it comprises, hence making complex space travel problems easier to manage.

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Most popular questions from this chapter

A hot-air balloonist, rising vertically with a constant velocity of magnitude \(5.00 \mathrm{m} / \mathrm{s},\) releases a sandbag at an instant when the balloon is 40.0 \(\mathrm{m}\) above the ground (Fig. 2.41\()\) . After it is released, the sandbag is in frec fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds s strike after its release will the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch \(a_{y}-t\) \(v_{y^{-} t},\) and \(y-t\) graphs for the motion.

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help." When she has fallen for 3.0 \(\mathrm{s}\) , she hears the echo of her shout from the valley floor below. The speed of sound is 340 \(\mathrm{m} / \mathrm{s}\) . (a) How tall is the cliff? \((b)\) If air resistance is neglected, bow fast will she be moving just before she hits the ground? (Her actual speed will be less than this, due to air resistance.)

Large cockroaches can run as fast as 1.50 \(\mathrm{m} / \mathrm{s}\) in short bursts. Suppose you turn on the light in a cheap motel and see one scurrying directly away from you at a constant 1.50 \(\mathrm{m} / \mathrm{s}\) . If you start 0.90 \(\mathrm{m}\) behind the cockroach with an initial speed of 0.80 \(\mathrm{m} / \mathrm{s}\) toward it, what minimum constant acceleration would you need to catch up with it when it has traveled \(1.20 \mathrm{m},\) just short of safety under a counter?

A world-class sprinter accelerates to his maximum speed in 4.0 \(\mathrm{s}\) . He then maintains this speed for the remainder of a \(100-\mathrm{m}\) race, finishing with a total time of 9.1 \(\mathrm{s}\) . (a) What is the runner's average acceleration during the first 4.0 \(\mathrm{s} ?(\mathrm{b})\) What is his average acceleration during the last 5.1 \(\mathrm{s} ?(\mathrm{c})\) What is his average acceleration for the entire race? (d) Explain why your answer to part (c) is not the average of the answers to parts (a) and (b).

You are standing at rest at a bus stop. A bus moving at a constant speed of 5.00 m/s passes you. When the rear of the bus is 12.0 m past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of 0.960 m/s. How far would you have to run before you catch up with the rear of the bus, and how fast must you be running then? Would an average college student be physically able to accomplish this?

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