91Ó°ÊÓ

Mars Landing. In January \(2004,\) NASA landed exploration vehicles on Mars. Part of the descent consisted of the following stages: Stage \(A:\) Friction with the atmosphere reduced the speed from \(19,300 \mathrm{km} / \mathrm{h}\) to 1600 \(\mathrm{km} / \mathrm{h}\) in 4.0 \(\mathrm{min.}\) Stage \(B:\) A parachute then opened to slow it down to 321 \(\mathrm{km} / \mathrm{h}\) in 94 \(\mathrm{s} .\) Stage \(C :\) Retro rockets then fired to reduce its speed to zero over a distance of 75 \(\mathrm{m}.\) Assume that each stage followed immediately after the preceding one and that the acceleration during each stage was constant. (a) Find the rocket's acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ) during each stage. (b) What total distance (in km) did the rocket travel during stages \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C} ?\)

Step by step solution

01

Convert Units for Stage A

First, convert the initial and final speeds from km/h to m/s. \(v_i = 19300 \, \text{km/h} = \frac{19300 \times 1000}{3600} \, \text{m/s}\) and \(v_f = 1600 \, \text{km/h} = \frac{1600 \times 1000}{3600} \, \text{m/s}\). The time duration is 4.0 min, which is \(4 \times 60 = 240\, \text{s}\).

02

Calculate Acceleration for Stage A

Use the formula \(a = \frac{v_f - v_i}{t}\) to compute the acceleration. After substituting values, we get \(a = \frac{1600 \times \frac{1000}{3600} - 19300 \times \frac{1000}{3600}}{240}\). Evaluate the expression to find the acceleration for Stage A.

03

Convert Units for Stage B

Convert the speed values from km/h to m/s for Stage B: \(v_i = 1600 \, \text{km/h} = \frac{1600 \times 1000}{3600} \, \text{m/s}\) and \(v_f = 321 \, \text{km/h} = \frac{321 \times 1000}{3600} \, \text{m/s}\). The time duration is 94 s.

04

Calculate Acceleration for Stage B

Apply the formula \(a = \frac{v_f - v_i}{t}\) with the converted values. \(a = \frac{321 \times \frac{1000}{3600} - 1600 \times \frac{1000}{3600}}{94}\). Calculate the result to find the acceleration for Stage B.

05

Calculate Acceleration for Stage C

Start with the initial speed \(v_i = 321 \, \text{km/h} = \frac{321 \times 1000}{3600} \, \text{m/s}\), and final speed \(v_f = 0\). Use the formula \(v_f^2 = v_i^2 + 2a d\), where \(d = 75 \, \text{m}\), to find \(a\). Rearranging gives \(a = \frac{-v_i^2}{2d}\). Substitute the converted initial velocity to solve.

06

Determine Distance for Stage A

Use the kinematic equation \(d = v_i t + \frac{1}{2}a t^2\) with Stage A's values. Compute using the acceleration from Step 2 to find distance \(d_A\).

07

Determine Distance for Stage B

Apply the distance formula \(d = v_i t + \frac{1}{2}a t^2\) using Stage B's numbers. Calculate with the acceleration from Step 4 to find distance \(d_B\).

08

Calculate Total Distance for Stages A, B, and C

Add distances from each stage: \(\text{Total distance} = d_A + d_B + d_C\). Convert to kilometers as needed: \(1 \, \text{m} = 0.001 \, \text{km}\).

09

Final Results

Calculate the accelerations from Steps 2, 4, and 5. Then, the total distance from Step 8 gives the answer for part (b).

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Key Concepts

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

Mars landing

Mars landings are complex events that involve multiple stages to ensure a safe descent. NASA's exploration vehicles, like the rovers in 2004, undergo different stages as they slow down from the high speeds attained while entering the Martian atmosphere.
In 2004, the Mars exploration vehicles experienced three descent stages: atmospheric friction, parachute deployment, and retro rocket firing. Each stage has a critical role. Together, they ensure a controlled landing on Mars's surface.
Understanding these stages involves physics principles like kinematics and dynamics. During each stage, constant acceleration is assumed, which simplifies the calculations needed to evaluate how these craft slow down and ultimately stop safely on Mars.

Acceleration calculation

Acceleration is critical in understanding how an object changes speed. It describes how quickly velocity changes per unit of time. In the Mars landing problem, each descent stage has a distinct acceleration value, which must be calculated separately.
To calculate acceleration, we use the formula: \[ a = \frac{v_f - v_i}{t} \]where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, and \(t\) is the time the change in velocity occurs over.
This fundamental formula applies to both Stage A and B of the Mars landing. For Stage C, another formula from kinematic equations is used, given in the form of:\[ v_f^2 = v_i^2 + 2a d \]Here, rearranging for \(a\) and solving allows us to determine the acceleration when the distance is known.

Kinematic equations

Kinematic equations play a pivotal role in solving physics problems related to motion, such as the Mars landing exercise. These equations help us track details like velocity changes and the distance covered over time intervals when accelerations are constant.
Key kinematic equations include:

• The velocity equation: \(v_f = v_i + at\)
• The distance formula with initial and final speeds: \(d = v_i t + \frac{1}{2}at^2\)
• The equation relating velocity squares and acceleration over distance: \(v_f^2 = v_i^2 + 2ad\)

In our Mars landing problem, these equations allow us to derive accelerations and distances for each stage during the descent.
Understanding and correctly applying kinematic equations are essential for solving real-world physics problems, illustrating how theoretical physics is used in practical applications.