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A rocket carrying a satellite is accelerating straight up from the earth's surface. At \(1.15 \mathrm{~s}\) 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 \(5.90 \mathrm{~s}\) of its flight.

Short Answer

Expert verified
The average velocity of the rocket for the 4.75 s portion of the flight is approximately \(197.26 \mathrm{m/s}\), while the average velocity for the full 5.90 s flight is approximately \(169.49 \mathrm{m/s}\).

Step by step solution

01

Determine Average Velocity for the First Section of the Flight

Find the average velocity from \(63 \mathrm{~m}\) to \(1.00 \mathrm{~km}\) over \(4.75 \mathrm{~s}\). The distance covered is given by \((1.00 \mathrm{km}- 63 \mathrm{m})\), and the time taken is \(4.75 \mathrm{s}\). The displacement is \(1.00 \mathrm{km} - 63 \mathrm{m} = 937 \mathrm{m}\). Therefore, the average velocity for this part of the flight is \( \frac{937 \mathrm{m}}{4.75 \mathrm{s}} \).
02

Determine Average Velocity for the Full Flight

Next, we need to find the average velocity from \(0 \mathrm{~m}\) (launch point) to \(1.00 \mathrm{~km}\) over a total time of \(5.90 \mathrm{~s}\). Hence, the displacement here is \(1.00 \mathrm{km} = 1000 \mathrm{m}\). Hence, the average velocity for the whole flight is: \( \frac{1000 \mathrm{m}}{5.90 \mathrm{s}} \).

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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 topic in physics that involves analyzing how rockets move through space or from the ground up. When a rocket launches, it accelerates upward against the force of gravity. This requires a powerful boost of force from the engines and carefully controlled motion.
A rocket must navigate through several stages of motion, characterized by initial acceleration, main flight, and reaching final speed. Each of these stages involves different physics principles, such as thrust, drag, and gravitational pull. Understanding rocket motion helps us appreciate the complexity of space travel and its engineering marvels.
  • Initial lift-off: the vehicle overcomes Earth's gravity.
  • Main stage: propulsion maintains speed or increases velocity.
  • Final phase: reaching orbit or desired altitude.
The complexity of rocket motion arises from the necessity to balance various forces and maintain stability. Engineers must account for these factors to ensure a successful mission. This makes studying rocket motion intriguing yet challenging.
Kinematics
Kinematics is the branch of physics that describes how objects move, focusing primarily on their velocity, speed, displacement, and time. It does not consider the causes of motion, such as forces or energy, but is crucial for understanding movement patterns.
In rocket motion, kinematics helps us predict the path a rocket will take given certain initial conditions. It helps calculate quantities like average velocity, which is key when examining a rocket's journey. Average velocity refers to the total displacement divided by the total time taken.
  • Displacement: total change in position.
  • Velocity: speed in a specified direction.
  • Time: period over which motion occurs.
When solving problems related to rocket kinematics, such as the one in the exercise, it is essential to break down the motion into specific time intervals or distances. This way, you can easily determine the velocity and understand the dynamics of the rocket's path.
Displacement Calculation
Displacement calculation is an integral part of solving kinematics problems. Displacement refers to the change in position from an initial point to a final point. It is a vector quantity, meaning it has both magnitude and direction.
To calculate displacement for a rocket, you determine the straight-line distance between its starting position and its final position. For a problem like the one in the exercise, you need to clearly identify these positions and subtract to find the net change.The displacement formula used in the exercise:\[ \text{Displacement} = \text{Final Position} - \text{Initial Position} \]For example, figuring out the rocket's displacement from clearing the platform to reaching 1.00 km above the ground involves taking the final position and subtracting the starting position just above the platform. Hence, calculating 1.00 km and subtracting 63 m gives a displacement of 937 m over the 4.75 s.
Correctly calculating displacement is crucial for determining other aspects such as average velocity, which provides insights into the overall speed and efficiency of the rocket's movement.

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Most popular questions from this chapter

In the fastest measured tennis serve, the ball left the racquet at \(73.14 \mathrm{~m} / \mathrm{s}\). A served tennis ball is typically in contact with the racquet for \(30.0 \mathrm{~ms}\) and starts from rest. Assume constant acceleration. (a) What was the ball's acceleration during this serve? (b) How far did the ball travel during the serve?

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If the aorta (diameter \(d_{\mathrm{a}}\) ) branches into two equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches? (a) \(\sqrt{d_{a}} ;\) (b) \(d_{\mathrm{a}} / \sqrt{2} ;\) (c) \(2 d_{\mathrm{a}} ;\) (d) \(d_{\mathrm{a}} / 2\)

The acceleration of a bus is given by \(a_{x}(t)=\alpha t,\) where \(\alpha=1.2 \mathrm{~m} / \mathrm{s}^{3} .\) (a) If the bus's velocity at time \(t=1.0 \mathrm{~s}\) is \(5.0 \mathrm{~m} / \mathrm{s}\) what is its velocity at time \(t=2.0 \mathrm{~s} ?\) (b) If the bus's position at time \(t=1.0 \mathrm{~s}\) is \(6.0 \mathrm{~m},\) what is its position at time \(t=2.0 \mathrm{~s} ?(\mathrm{c})\) Sketch \(a_{y}-t\) \(v_{y}-t,\) and \(x-t\) graphs for the motion.

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