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Chapter 3: Problem 3

\(A\) "sun yacht" is a spacecraft with a large sail that is pushed by sunlight. Although such a push is tiny in everyday circumstances, it can be large enough to send the spacecraft outward from the Sun on a cost-free but slow trip. Suppose that the spacecraft has a mass of \(900 \mathrm{~kg}\) and reccives a push of \(20 \mathrm{~N}\). (a) What is the magnitude of the resulting acceleration? If the craft starts from rest, (b) how far will it travel in 1 day and (c) how fast will it then be moving?

Short Answer

Expert verified
Acceleration is 0.0222 m/s虏, distance traveled is 82,858,656 meters, and final velocity is 1,917.6 m/s.

Step by step solution

01

Identify given values

Mass of the spacecraft, m = 900 kg. Force received by spacecraft, F = 20 N.
02

Calculate Acceleration

Use Newton's second law of motion, which states that the acceleration, a, can be calculated using the formula: \[ a = \frac{F}{m} \] Substitute the given values: \[ a = \frac{20 \mathrm{~N}}{900 \mathrm{~kg}} = 0.0222 \mathrm{~m/s^2} \]
03

Determine time period

Convert 1 day into seconds to use in our calculations. \[ 1 \text{ day} = 24 \times 60 \times 60 = 86,400 \text{ seconds} \]
04

Calculate distance traveled

Since the spacecraft starts from rest, we use the formula for distance traveled under constant acceleration: \[ s = ut + \frac{1}{2}at^2 \] Where initial velocity, u = 0, time t = 86,400 seconds, and acceleration a = 0.0222 m/s虏. \[ s = 0 + \frac{1}{2} \times 0.0222 \mathrm{~m/s^2} \times (86,400 \mathrm{~s})^2 \] \[ s = 0.0111 \times 7,464,960,000 \] \[ s = 82,858,656 \text{ meters} \]
05

Calculate final velocity

Use the final velocity formula for constant acceleration: \[ v = u + at \] Since initial velocity, u = 0, acceleration a = 0.0222 m/s虏, and time t = 86,400 seconds: \[ v = 0 + 0.0222 \times 86,400 \] \[ v = 1,917.6 \text{ m/s} \]

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Key Concepts

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

acceleration calculation
Acceleration is a measure of how quickly an object changes its velocity. It鈥檚 calculated by the formula from Newton鈥檚 second law of motion:
\( a = \frac{F}{m} \)
Here, \( F \) represents the force applied to the object, and \( m \) is the mass of the object. By substituting the given values, we have:
\( a = \frac{20 \mathrm{~N}}{900 \mathrm{~kg}} = 0.0222 \mathrm{~m/s^2} \)
This means that the spacecraft will experience an acceleration of \( 0.0222 \mathrm{~m/s^2} \), which is due to the force exerted by sunlight on the solar sail. Even though the force of 20 N seems tiny, given the spacecraft鈥檚 mass, the resultant acceleration will gradually increase its speed over time.
constant acceleration formulas
When an object is subjected to a constant acceleration, several formulas can be used to determine its motion characteristics. These formulas are derived from the basic principles of kinematics. Some important ones include:
\( s = ut + \frac{1}{2}at^2 \)
\( v = u + at \)
where \( u \) is the initial velocity, \( a \) is the constant acceleration, \( t \) is the time, and \( s \) is the distance the object has traveled. These formulas help us calculate the distance traveled by the spacecraft after a given time and its final speed.
distance traveled under acceleration
To find the distance traveled by the spacecraft, use the formula:
\( s = ut + \frac{1}{2}at^2 \)
Given that the spacecraft starts from rest \( (u = 0) \), the formula simplifies to:
\( s = \frac{1}{2}at^2 \)
Substituting the given values: \( a = 0.0222 \mathrm{~m/s^2} \) and \( t = 86,400 \mathrm{~seconds} \), we get:
\( s = \frac{1}{2} \times 0.0222 \mathrm{~m/s^2} \times (86,400 \mathrm{~s})^2 \)
\( s = 0.0111 \times 7,464,960,000 = 82,858,656 \mathrm{~meters} \)
So, the spacecraft travels a distance of 82,858,656 meters (or roughly 82,859 km) in one day.
final velocity formula
The final velocity of an object under constant acceleration can be calculated using:
\( v = u + at \)
Since the spacecraft starts from rest \( (u = 0) \), this simplifies to:
\( v = at \)
Substituting the acceleration and time values: \( a = 0.0222 \mathrm{~m/s^2} \) and \( t = 86,400 \mathrm{~seconds} \), we have:
\( v = 0.0222 \times 86,400 = 1,917.6 \mathrm{~m/s} \)
Therefore, the spacecraft's final velocity after one day is 1,917.6 meters per second (or about 1.92 km/s). This shows how a small constant force can significantly accelerate an object over time.

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