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Chapter 5: Problem 71

IP While waiting at the airport for your flight to leave, you observe some of the jets as they take off. With your watch you find that it takes about 35 seconds for a plane to go from rest to takeoff speed. In addition, you estimate that the distance required is about \(1.5 \mathrm{km}\). (a) If the mass of a jet is \(1.70 \times 10^{5} \mathrm{kg}\) what force is needed for takeoff? (b) Describe the strategy you used to solve part (a).

Short Answer

Expert verified
The force needed for takeoff is \( 4.165 \times 10^5 \, \mathrm{N} \).

Step by step solution

01

Identify the known values

The time for takeoff is given as 35 seconds. The distance for takeoff is 1.5 km, which is converted to meters as 1500 meters. The mass of the jet is given as \(1.70 \times 10^{5} \mathrm{kg}\).
02

Use the kinematic formula

To find the force, we first calculate acceleration. For this, we use the kinematic equation \( s = ut + \frac{1}{2}at^2 \), where \(s\) is the distance, \(u\) is the initial velocity (0 in this case), \(a\) is acceleration, and \(t\) is time. Plug in the values: \(1500 = 0 \cdot 35 + \frac{1}{2}a \cdot 35^2 \).
03

Solve for acceleration

Simplify and solve the equation for acceleration: \[1500 = \frac{1}{2}a \cdot 1175\]Multiply both sides by 2 and divide by 1225 to find \(a\) : \[ a = \frac{3000}{1225} a \approx 2.45 \, \mathrm{m/s^2}\]
04

Use Newton's Second Law to find force

Now that we have the acceleration, use Newton's Second Law \( F = ma \) to find the force. Substitute the values: \[F = 1.70 \times 10^5 \cdot 2.45\]Calculate to find \(F\): \[F = 4.165 \times 10^5 \, \mathrm{N}\]
05

Describe the strategy

The strategy involves first calculating the acceleration using the kinematic equation with the given distance and time. With the acceleration found, Newton's Second Law, \( F = ma \), is then applied to find the force required for takeoff.

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Key Concepts

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

Kinematic Equations
Kinematic equations are useful tools for understanding motion in physics, especially when dealing with uniformly accelerated motion. In the given problem, we observe a jet taking off, and we know the time it takes to reach takeoff speed and the distance it covers. To find the acceleration, we use the kinematic equation:\[ s = ut + \frac{1}{2}at^2 \]In this equation:
  • \( s \) represents the displacement or distance traveled, which is 1500 meters in this scenario.
  • \( u \) is the initial velocity. Since the jet starts from rest, \( u = 0 \).
  • \( a \) represents acceleration, which we need to find.
  • \( t \) is the time, given as 35 seconds.
To solve for the acceleration \( a \), substitute the known values into the equation, then simplify and solve for \( a \). This process helps to understand how objects move over time under constant acceleration.
Acceleration Calculation
Acceleration is a key concept for determining how an object's velocity changes over time. In our airport observation problem, we solved for acceleration using the kinematic equation:\[1500 = \frac{1}{2}a \cdot 35^2\]To isolate \( a \), the following steps are performed:
  • First, multiply both sides by 2 to eliminate the fraction: \( 3000 = a \cdot 1225 \)
  • Next, divide both sides by 1225 to solve for \( a \): \( a = \frac{3000}{1225} \approx 2.45 \ \mathrm{m/s^2} \)
It's important to understand that acceleration is the rate of change of velocity and is calculated as meters per second squared (m/s虏). With the acceleration known, we can move on to finding other important dynamics, like the force needed for jet takeoff.
Force Calculation
Once we determine the acceleration, Newton's Second Law allows us to calculate the force needed for the jet's takeoff. Newton's Second Law is represented by the equation:\[ F = ma \]Where:
  • \( F \) is the force in newtons needed to accelerate the jet.
  • \( m \) is the mass of the jet, given as \(1.70 \times 10^5 \ \mathrm{kg}\).
  • \( a \) is the acceleration we previously calculated, approximately \(2.45 \ \mathrm{m/s^2}\).
By substituting the values into the equation, we find:\[ F = 1.70 \times 10^5 \times 2.45 \approx 4.165 \times 10^5 \ \mathrm{N} \]This tells us that a force of about 416,500 newtons is required for the jet to reach takeoff speed in the given time and distance. Understanding how force combines with mass and acceleration through Newton's Second Law is fundamental in physics, and essential for studying how vehicles and objects move.

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Most popular questions from this chapter

When you weigh yourself on good old terra firma (solid ground), your weight is 142 lb. In an elevator your apparent weight is 121 lb. What are the direction and magnitude of the elevator's acceleration?

A 51.5 -kg swimmer with an initial speed of \(1.25 \mathrm{m} / \mathrm{s}\) decides to coast until she comes to rest. If she slows with constant acceleration and stops after coasting \(2.20 \mathrm{m},\) what was the force exerted on her by the water?

A \(5.0-\mathrm{kg}\) bag of potatoes sits on the bottom of a stationary shopping cart. (a) Sketch a free-body diagram for the bag of potatoes. (b) Now suppose the cart moves with a constant velocity. How does this affect your free-body diagram? Explain.

\- ce Predict/Explain You drop two balls of equal diameter from the same height at the same time. Ball 1 is made of metal and has a greater mass than ball \(2,\) which is made of wood. The upward force due to air resistance is the same for both balls. (a) Is the drop time of ball 1 greater than, less than, or equal to the drop time of ball \(2 ?\) (b) Choose the best explanation from among the following: I. The acceleration of gravity is the same for all objects, regardless of mass. II. The more massive ball is harder to accelerate. III. Air resistance has less effect on the more massive ball.

IP The VASIMR Rocket NASA plans to use a new type of rocket, a Variable Specific Impulse Magnetoplasma Rocket (VASIMR), on future missions. A VASIMR can produce \(1200 \mathrm{N}\) of thrust (force) when in operation. If a VASIMR has a mass of \(2.2 \times 10^{5} \mathrm{kg},\) (a) what acceleration will it experience? Assume that the only force acting on the rocket is its own thrust, and that the mass of the rocket is constant. (b) Over what distance must the rocket accelerate from rest to achieve a speed of \(9500 \mathrm{m} / \mathrm{s} ?\) (c) When the rocket has covered one-quarter the acceleration distance found in part (b), is its average speed \(1 / 2,\) \(1 / 3,\) or \(1 / 4\) its average speed during the final three-quarters of the acceleration distance? Explain.
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