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Chapter 2: Problem 36

An airplane travels 280 \(\mathrm{m}\) down the runway before taking off. Assuming that it has constant acceleration, if it starts from rest and becomes airborne in 8.00 s, how fast \((\) in \(\mathrm{m} / \mathrm{s})\) is it moving at takeoff?

Short Answer

Expert verified
The airplane's speed at takeoff is 70 m/s.

Step by step solution

01

Identify known variables

We know the airplane travels 280 meters (displacement \(d = 280 \, \text{m}\)), starts from rest (initial velocity \(v_i = 0 \, \text{m/s}\)), and becomes airborne in 8 seconds (time \(t = 8 \, \text{s}\)). We need to find the final velocity \(v_f\) at takeoff.
02

Use the kinematic equation

We use the kinematic equation: \[ d = v_i t + \frac{1}{2} a t^2 \] Since the initial velocity \(v_i = 0\), the equation simplifies to:\[ d = \frac{1}{2} a t^2 \] Rearranging for acceleration \(a\), we get:\[ a = \frac{2d}{t^2} \]
03

Calculate the acceleration

Substitute the known values into the equation for the acceleration:\[ a = \frac{2 \times 280}{8^2} = \frac{560}{64} = 8.75 \, \text{m/s}^2 \] The acceleration of the airplane is \(8.75 \, \text{m/s}^2\).
04

Find the final velocity

Use the kinematic equation for final velocity:\[ v_f = v_i + a t \] Substitute the values (\(v_i = 0\), \(a = 8.75 \, \text{m/s}^2\), \(t = 8 \, \text{s}\)):\[ v_f = 0 + 8.75 \times 8 = 70 \, \text{m/s} \] Hence, the airplane's speed at takeoff is \(70 \, \text{m/s}\).

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Key Concepts

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

Constant Acceleration
In physics, constant acceleration refers to a scenario where the acceleration of an object remains the same throughout the entire duration of its motion. This can be observed when an object speeds up or slows down at a steady rate. Understanding constant acceleration is crucial since it is a simplifying assumption that makes many kinematic problems easier to solve.

Here鈥檚 what constant acceleration means in a real-world example, like the airplane on a runway:
  • The plane starts from rest, which means its initial velocity is zero.
  • It accelerates at a steady rate until it achieves the speed necessary for takeoff.
  • In this scenario, the plane's motion can be accurately described by kinematic equations, thanks to its constant acceleration.
Recognizing constant acceleration allows us to use these equations to predict an object's velocity at any given time, its displacement, and overall motion behavior.
Kinematic Equations
Kinematic equations are a set of mathematical formulas used in physics to predict the future position and velocity of an object undergoing uniform acceleration. They are incredibly useful in solving motion-related problems such as our airplane exercise.

Let's break down how one of these equations helps in our case:
  • We used the equation \( d = v_i t + \frac{1}{2} a t^2 \) which simplifies to \( d = \frac{1}{2} a t^2 \) when the initial velocity \( v_i \) is zero.
  • This equation connects displacement \( d \), acceleration \( a \), and time \( t \) allowing us to solve for acceleration given the other two variables.
  • Additionally, the equation \( v_f = v_i + a t \) helps find the final velocity after a period of time.

    Understanding how to rearrange and apply these equations can be very powerful in physics, as they allow you to precisely calculate different aspects of motion like displacement, time, and velocity under constant acceleration conditions.
Final Velocity
Final velocity is an important concept in kinematics, representing the velocity an object has at the end of a given period of time under acceleration. In the context of our exercise, it answers the question of how fast the airplane is moving right at the moment of takeoff.

Here is how to think about final velocity:
  • It marks the point of transition from one phase of motion to another (e.g., from rest to takeoff).
  • Using the kinematic equation \( v_f = v_i + a t \), you can determine the final velocity by considering initial velocity, acceleration, and time.
  • In our scenario, the initial velocity \( v_i \) is zero because the plane starts at rest, and thus the final velocity can be directly computed from product of acceleration and time.
Understanding the final velocity not only tells us how fast an object is moving at a specific instance but also provides insights into the object's overall dynamic behavior in situations where acceleration is constant.

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Most popular questions from this chapter

In \(1954,\) Roger Bannister became the first human to run a mile in less than 4 minutes. Suppose that a runner on a straight track covers a distance of 1.00 mi in exactly 4.00 min. What is his average speed in (a) \(\mathrm{mi} / \mathrm{h},\) (b) \(\mathrm{ft} / \mathrm{s},\) and \((\mathrm{c}) \mathrm{m} / \mathrm{s}\) ?

A car driving on the turnpike accelerates uniformly in a straight line from 88 \(\mathrm{ft} / \mathrm{s}\) (60 mph) to 110 \(\mathrm{ft} / \mathrm{s}\) (75 mph) in 3.50 \(\mathrm{s} .\) (a) What is the car's acceleration? (b) How far does the car travel while it accelerates?

At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a constant acceleration of 2.50 \(\mathrm{m} / \mathrm{s}^{2} .\) At the same instant, a truck, traveling with a constant speed of \(15.0 \mathrm{m} / \mathrm{s},\) overtakes and passes the automobile. (a) How far beyond its starting point does the automobile overtake the truck? (b) How fast is the automobile traveling when it overtakes the truck?

Two cars having equal speeds hit their brakes at the same time, but car \(A\) has three times the acceleration as car \(B .\) (a) If car A travels a distance \(D\) before stopping, how far (in terms of \(D )\) will car \(B\) go before stopping? (b) If car \(B\) stops in time \(T,\) how long (in terms of \(T )\) will it take for car \(A\) to stop?

How high is the cliff? Suppose you are climbing in the High Sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of this cliff, you drop a rock from the top and, 10.0 s later, hear the sound of it hitting the ground at the foot of the cliff. (a) Ignoring air resistance, how high is the cliff if the speed of sound is 330 \(\mathrm{m} / \mathrm{s} ?\) (b) Suppose you had ignored the time it takes the sound to reach you. In that case, would you have overestimated or underestimated the height of the cliff? Explain your reasoning.
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