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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 \(\mathrm{s}\) 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 (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 \(\mathrm{min} 7\) (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 \(\mathrm{s}\) to 1.00 \(\mathrm{min}\) ?

Step by step solution

01

Convert Initial and Final Velocities to m/s

First, we need to convert the initial and final velocities from km/h to m/s. For the first interval (0 to 8 s):- Initial velocity (\(v_0\)) = 0 km/h = 0 m/s- Final velocity (\(v_{f1}\)) = 161 km/h = \(\frac{161 \times 1000}{3600} = 44.72\) m/s.For the second interval (8 s to 1 min):- Initial velocity = 161 km/h = 44.72 m/s (velocity at 8 s)- Final velocity (\(v_{f2}\)) = 1610 km/h = \(\frac{1610 \times 1000}{3600} = 447.22\) m/s.

02

Calculate Average Acceleration for First Interval

Use the formula for average acceleration: \[ a = \frac{v_f - v_0}{t} \]For the first interval, from rest (0 m/s) to 44.72 m/s in 8 s:\[ a_1 = \frac{44.72 - 0}{8} = 5.59 \text{ m/s}^2 \]

03

Calculate Average Acceleration for Second Interval

For the second interval, from 8 s to 1 minute (60 s), calculate the average acceleration using:\[ a_2 = \frac{v_{f2} - v_{f1}}{t} \]where \(t = 60 - 8 = 52\) s:\[ a_2 = \frac{447.22 - 44.72}{52} = 7.74 \text{ m/s}^2 \]

04

Calculate Distance Traveled During First Interval

For the first 8 s with constant acceleration, use the equation:\[ d = v_0 t + \frac{1}{2} a t^2 \]Substituting initial velocity (0 m/s), acceleration (5.59 m/s²), and time (8 s):\[ d_1 = 0 \times 8 + \frac{1}{2} \times 5.59 \times 8^2 = 178.88 \text{ meters} \]

05

Calculate Distance Traveled During Second Interval

For the interval from 8 s to 60 s, again use the distance formula:\[ d = v_{i} t + \frac{1}{2} a t^2 \]where initial velocity \(v_i = 44.72\) m/s, acceleration \(a_2 = 7.74\) m/s², and \(t = 52\) s:\[ d_2 = 44.72 \times 52 + \frac{1}{2} \times 7.74 \times 52^2 = 1535.44 + 1045.68 = 2581.12 \text{ meters} \]

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Key Concepts

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

Average Acceleration

To understand how fast an object's speed changes over time, average acceleration is key. It's like figuring out how quickly a car speeds up when it moves away from a stop sign. For a spacecraft, like the Space Shuttle at launch, average acceleration shows how it picks up speed. It's calculated using the formula:

• \( a = \frac{v_f - v_0}{t} \)

Here, \(v_f\) is the final speed, \(v_0\) is the initial speed, and \(t\) is the time taken. During the first 8 seconds of launch, the Shuttle's average acceleration was from 0 m/s to 44.72 m/s, giving an acceleration of 5.59 m/s². In the next phase, over 52 seconds, it accelerated from 44.72 m/s to 447.22 m/s, resulting in an average acceleration of 7.74 m/s².

Conversion of Units

Before you can apply physics formulas correctly, you might need to convert units, especially when dealing with velocities. For space flights and similar exercises, velocity is often given in kilometers per hour (km/h). But, to use it in equations for acceleration or distance, it's typically changed to meters per second (m/s).

• To convert km/h to m/s, the formula is:\[ \text{velocity in m/s} = \frac{\text{velocity in km/h} \times 1000}{3600} \]

For example, converting 161 km/h results in approximately 44.72 m/s. This simple but crucial step ensures the calculations match the standard units in physics problems. Similarly, 1610 km/h turns into 447.22 m/s, providing an accurate value for further calculations.

Constant Acceleration

Constant acceleration refers to a steady change in velocity over time. Think of spreading peanut butter evenly on bread - acceleration spreads evenly across time intervals. In physics problems like this one, it's assumed that acceleration within defined intervals doesn't change, simplifying calculations.With constant acceleration:

• Any speed increase happens uniformly over the time interval.
• Equations like \( v = v_0 + at \) and \( d = v_0t + \frac{1}{2}at^2 \) are used.

For the Shuttle, during each interval, constant acceleration was assumed separately, which means even though the overall acceleration wasn't steady, within each interval it was calculated as if it was. This makes problem-solving more manageable and avoids complex variable accelerations.

Distance Calculation

Finding out how far something travels when it's speeding up can be tricky, but with acceleration, we have the tools to do it! Using the formula for distance,

• \( d = v_0t + \frac{1}{2}at^2 \)

we can calculate the distance traveled under constant acceleration. First, during the initial 8 seconds when the Space Shuttle began from rest, the distance was 178.88 meters.
In the next segment from 8 seconds to 1 minute, the Shuttle covered 2581.12 meters. By adding these numbers, we understand how far the Shuttle moved at varying speeds - a combination of its initial velocity and how quickly it accelerated over time.