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Chapter 11: Problem 33

An airplane begins its take-off run at \(A\) with zero velocity and a constant acceleration \(a\). Knowing that it becomes airborne 30 s later at \(B\) and that the distance \(A B\) is \(900 \mathrm{m}\), determine ( \(a\) ) the acceleration \(a\) (b) the take-off velocity \(v_{B}\).

Short Answer

Expert verified
The acceleration \( a \) is 2 m/s虏, and the take-off velocity \( v_B \) is 60 m/s.

Step by step solution

01

Identify the Known Variables

We know the initial velocity \( v_A = 0 \) m/s, the time to become airborne \( t = 30 \) seconds, and the distance \( s = 900 \) meters.
02

Use the Kinematic Equation

Use the equation for constant acceleration: \[ s = v_A t + \frac{1}{2} a t^2 \]. We can plug in the values: \[ 900 = 0 \cdot 30 + \frac{1}{2} a (30)^2 \].
03

Solve for the Acceleration \( a \)

Simplify the equation: \[ 900 = 0 + \frac{1}{2} a \cdot 900 \]. This simplifies to \[ 900 = 450a \]. Solve for \( a \) by dividing both sides by 450: \[ a = \frac{900}{450} = 2 \text{ m/s}^2 \].
04

Calculate the Take-off Velocity \( v_B \)

Use the formula \( v = v_A + at \). Since the initial velocity \( v_A = 0 \), it simplifies to \( v_B = 0 + 2 \cdot 30 \). Calculate \( v_B = 60 \text{ m/s} \).

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Key Concepts

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

Constant Acceleration
Constant acceleration is a fundamental concept in kinematics that refers to a situation where an object's velocity changes at a uniform rate over time. This means that the object is increasing or decreasing its speed by the same amount every second.
Understanding constant acceleration can help solve problems involving moving objects, such as the take-off of an airplane. When acceleration is constant, it greatly simplifies calculations, allowing us to predict future motion accurately. Here's what you need to know:
  • Acceleration (\( a \)) remains the same throughout the motion.
  • Allows us to use straightforward kinematic equations to find unknowns.
  • Helps in determining velocities and displacements over time.
In the case of an airplane starting from rest, like in the exercise you've seen, constant acceleration means the airplane consistently speeds up until takeoff.
Velocity
Velocity refers to the speed of something in a given direction. Unlike speed, which only tells you how fast an object is moving, velocity provides information about the direction of movement as well.
In the context of the airplane take-off exercise, velocity is crucial at two different points in time:
  • Initial velocity (\( v_A \)) which was 0 m/s because the airplane started from rest.
  • Final velocity (\( v_B \)) which you calculated using the formula \( v = v_A + at \).
This formula simply shows how the velocity changes over time with constant acceleration (\(a\)). The final velocity (\(v_B\)) turned out to be 60 m/s after 30 seconds.
Velocity helps to describe how and when the airplane becomes airborne, and it gives a full picture of its motion from start until take-off.
Kinematic Equation
Kinematic equations are a set of equations that express mathematically how objects in motion behave under constant acceleration. They allow us to find various unknown values, such as displacement, velocity, time, and acceleration.
In this particular exercise, the most important kinematic equation used was:\[ s = v_A t + \frac{1}{2} a t^2\]This equation allows us to calculate the displacement (\(s\)) when we know the time (\(t\)), initial velocity (\(v_A\)), and acceleration (\(a\)). Key points about kinematic equations:
  • Used when acceleration is constant.
  • Can be rearranged to solve for different unknown variables.
  • Highly useful in practical scenarios involving motion, such as vehicles, projectiles, and more.
By applying the kinematic equation, you were able to determine the acceleration needed for the airplane to cover 900 meters in 30 seconds. These equations are vital tools for any study of motion!

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Most popular questions from this chapter

Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit \(100 \mathrm{mi}\) above the surface of the planet. (See information given in Probs. \(11.153-11.154 .)\) $$ \begin{array}{l}{11.155 \text { Venus: } g=29.20 \mathrm{ft} / \mathrm{s}^{2}, R=3761 \mathrm{mi}} \\ {11.156 \mathrm{Mars}: g=12.17 \mathrm{ft} / \mathrm{s}^{2}, R=2102 \mathrm{mi}} \\ {11.157 \text { Jupiter: } g=75.35 \mathrm{ft} / \mathrm{s}^{2}, R=44,432 \mathrm{mi}}\end{array} $$

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