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Chapter 5: Problem 12

\(\bullet\) Stopping a 747 A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is \(3.50 \times 10^{5} \mathrm{kg}\) its speed is \(27.0 \mathrm{m} / \mathrm{s},\) and the net braking force is \(4.30 \times 10^{5} \mathrm{N}\) (a) what is its speed 7.50 s later? (b) How far has it traveled in this time?

Short Answer

Expert verified
(a) 17.8 m/s; (b) 168.0 m

Step by step solution

01

Calculate the acceleration

To determine the acceleration of the jetliner, use Newton's second law. The net force is given, and the mass of the jet is known. The equation is \( F = ma \), which can be rearranged to \( a = \frac{F}{m} \). Substituting in the given values, \( F = 4.30 \times 10^{5} \, \text{N} \) and \( m = 3.50 \times 10^{5} \, \text{kg} \), we get \( a = \frac{4.30 \times 10^{5}}{3.50 \times 10^{5}} = 1.23 \, \text{m/s}^2 \). The acceleration is actually negative because it is acting in the direction opposite to the motion.
02

Determine the final speed after 7.50 seconds

Use the equation \( v = u + at \) to calculate the final speed, where \( u = 27.0 \, \text{m/s} \) is the initial speed, \( a = -1.23 \, \text{m/s}^2 \) is the acceleration, and \( t = 7.50 \, \text{s} \) is the time. Substituting these values gives \( v = 27.0 + (-1.23) \times 7.50 \). This simplifies to \( v = 27.0 - 9.225 = 17.775 \, \text{m/s} \).
03

Calculate the distance traveled

To find the distance traveled, use the equation \( s = ut + \frac{1}{2}at^2 \). Substituting the known values \( u = 27.0 \, \text{m/s}, t = 7.50 \, \text{s}, \text{and } a = -1.23 \, \text{m/s}^2 \), we calculate \( s = 27.0 \times 7.50 + \frac{1}{2} \times (-1.23) \times (7.50)^2 \). This results in \( s = 202.5 - 34.46875 = 168.03125 \, \text{m} \).

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Key Concepts

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

Understanding Acceleration
Acceleration is a key concept rooted in Newton's Laws of Motion. It describes how quickly an object's velocity changes over time. In this exercise, the 747 jetliner decelerates because a braking force acts against its motion.
To find the acceleration, we use Newton's second law of motion, which states:
  • Force (\(F\)) = Mass (\(m\)) \(\times\) Acceleration (\(a\)).
Rearranging this equation gives us:
  • Acceleration (\(a\)) = Force (\(F\)) / Mass (\(m\)).
In the example, with a force of \(4.30 \times 10^5 \text{ N}\) and a mass of \(3.50 \times 10^5 \text{ kg}\), the calculated acceleration is \(1.23 \text{ m/s}^2\). However, since the plane is slowing down, this acceleration is negative, indicating deceleration.
Exploring Kinematics
Kinematics is the study of motion, fundamentally concerned with velocity, distance, and time. In our scenario, it helps us predict the jetliner's speed and distance over time.
To find how fast it's going after 7.50 seconds, we use:
  • Final velocity (\(v\)) = Initial velocity (\(u\)) + (Acceleration (\(a\)) \(\times\) Time (\(t\))).
Plugging in \(u = 27.0 \text{ m/s}\), \(a = -1.23 \text{ m/s}^2\), and \(t = 7.50 \text{ s}\), gives a final velocity (\(v\)) of \(17.775 \text{ m/s}\).
The distance traveled can also be calculated with:
  • Displacement (\(s\)) = (\(u\)\(\times\ \)t) + \(\frac{1}{2}\) \((a\ \times t^2)\).
Substituting in our known values again provides a distance of \(168.03 \text{ m}\) covered in 7.50 seconds.
The Role of Braking Force
Braking force is crucial in stopping motion, illustrating Newton's third law: For every action, there is an equal and opposite reaction. The braking force here is \(4.30 \times 10^5 \text{ N}\), acting in the opposite direction to the jetliner's motion.
This braking force is responsible for the negative acceleration seen in the problem. Because forces are vectors, direction is important 鈥� a negative acceleration means the force reduces the speed.
When calculating changes such as speed or distance during breaking, understanding this directional force is essential. It allows you to predict how and when the jetliner will come to a stop safely.

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Most popular questions from this chapter

On vacation, your \(1400-\mathrm{kg}\) car pulls a \(560-\mathrm{kg}\) trailer away from a stoplight with an acceleration of \(1.85 \mathrm{m} / \mathrm{s}^{2}\). (a) What is the net force exerted on the trailer? (b) What force does the trailer exert on the car? (c) What is the net force acting on the car?

A 92 -kg water skier floating in a lake is pulled from rest to a speed of \(12 \mathrm{m} / \mathrm{s}\) in a distance of \(25 \mathrm{m}\). What is the net force exerted on the skier, assuming his acceleration is constant?

Your groceries are in a bag with paper handles. The handles will tear off if a force greater than \(51.5 \mathrm{N}\) is applied to them. What is the greatest mass of groceries that can be lifted safely with this bag, given that the bag is raised (a) with constant speed, or (b) with an acceleration of \(1.25 \mathrm{m} / \mathrm{s}^{2} ?\)

Blo Gecko Feet Researchers have found that a gecko's foot is covered with hundreds of thousands of small hairs (setae) that allow it to walk up walls and even across ceilings. A single foot pad, which has an area of \(1.0 \mathrm{cm}^{2}\), can attach to a wall or ceiling with a force of \(11 \mathrm{N}\). (a) How many \(250-g\) geckos could be suspended from the ceiling by a single foot pad? (b) Estimate the force per square centimeter that your body exerts on the soles of your shoes, and compare with the \(11 \mathrm{N} / \mathrm{cm}^{2}\) of the sticky gecko foot.

When two people push in the same direction on an object of mass \(m\) they cause an acceleration of magnitude \(a_{1}\). When the same people push in opposite directions, the acceleration of the object has a magnitude \(a_{2}\). Determine the magnitude of the force exerted by each of the two people in terms of \(m, a_{1},\) and \(a_{2}\)
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