/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 An aeroplane is flying verticall... [FREE SOLUTION] | 91Ó°ÊÓ

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An aeroplane is flying vertically upwards. When it is at a height of \(1000 \mathrm{~m}\) above the ground and moving at a speed of \(367 \mathrm{~m} / \mathrm{s}\), a shot is fired at it with a speed of \(567 \mathrm{~m} \mathrm{~s}^{-1}\) from a point directly below it. What should be the acceleration of aeroplane so that it may escape from being hit? (1) \(>5 \mathrm{~ms}^{-2}\) (2) \(>10 \mathrm{~ms}^{-2}\) (3) \(<10 \mathrm{~ms}^{-2}\) (4) Not possible

Short Answer

Expert verified
The correct answer is (4) Not possible.

Step by step solution

01

Analyze the initial conditions

An airplane is initially at a height of 1000 m and moving upward with a speed of 367 m/s. A bullet is fired from the ground directly below with a speed of 567 m/s. We need to calculate if there exists an acceleration that will allow the airplane to keep avoiding the bullet.
02

Establish equations of motion

For both the airplane and the bullet, we will calculate their positions over time. The equation for the airplane's position is: \( x_{plane}(t) = 1000 + 367t + \frac{1}{2} a t^2 \), where \( a \) is the acceleration we want to find. For the bullet, the equation is: \( x_{bullet}(t) = 567t \).
03

Set conditions for escape

To escape being hit, the airplane's position function, \( x_{plane}(t) \), must never equal the bullet's position, \( x_{bullet}(t) \), for any time \( t > 0 \). To find this acceleration, solve the equation set by equating both position functions: \( 1000 + 367t + \frac{1}{2} at^2 = 567t \). Simplifying gives: \( \frac{1}{2} at^2 + (367 - 567)t + 1000 = 0 \).
04

Solve for acceleration

The equation is a quadratic in the form \( at^2 + bt + c = 0 \) with \( a = \frac{1}{2}a \), \( b = -200 \), \( c = 1000 \). For the plane to avoid the bullet, the discriminant \( b^2 - 4ac \) must be negative, avoiding real solutions for \( t \). Calculate: \( (-200)^2 - 2a \cdot 1000 < 0 \), leading to \( 40000 < 2000a \), so \( a > 20 \).
05

Compare with given options

The correct acceleration for the aeroplane to avoid being hit is greater than 20 \( \mathrm{ms}^{-2} \). Examine the options presented: (1) \( >5 \mathrm{~ms}^{-2} \), (2) \( >10 \mathrm{~ms}^{-2} \), (3) \( <10 \mathrm{~ms}^{-2} \), (4) Not possible. None of these match \( >20 \mathrm{~ms}^{-2} \). Hence, based on the options, it is reasonable to conclude escaping is not feasible with the given acceleration options.

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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 essential tools in physics to describe the motion of objects. They relate the motion variables—such as displacement, velocity, acceleration, and time—allowing us to calculate one when the others are known. In this scenario, we use kinematic equations to determine the motion of both the airplane and the bullet shot at it.
The standard form of kinematic equations includes:
  • \( v = u + at \)
  • \( s = ut + \frac{1}{2}at^2 \)
  • \( v^2 = u^2 + 2as \)
In the problem, the airplane's position over time is given by the equation: \[ x_{plane}(t) = 1000 + 367t + \frac{1}{2}a t^2 \] Here, \( 1000 \) m is the initial position, \( 367 \) m/s is the initial velocity, and \( a \) is the unknown acceleration. Meanwhile, the bullet's motion is described using a simpler equation since it moves upward from the ground without additional acceleration. Thus, \[ x_{bullet}(t) = 567t \] Understanding how to apply these equations in different scenarios allows us to predict the future positions of moving objects.
Quadratic Equation
A quadratic equation is a critical part of determining the possible meeting point of the bullet and the airplane. The standard form of a quadratic equation is: \[ ax^2 + bx + c = 0 \] Solving a quadratic equation helps us find the values of \( x \) that satisfy the equation. In our context, \( x \) represents the time \( t \) at which the positions of the airplane and bullet are equal.
We derive a quadratic equation by setting the position equations of the aircraft and bullet equal: \[ 1000 + 367t + \frac{1}{2} at^2 = 567t \] This simplifies into: \[ \frac{1}{2}at^2 + (367 - 567)t + 1000 = 0 \] Now, solving this yields the conditions under which the airplane could avoid being hit. The crucial point here is the discriminant of the quadratic equation:\[ b^2 - 4ac \] If it is negative, the quadratic equation has no real solutions for \( t \), meaning the airplane's position never coincides with the bullet's, ensuring safety.
Relative Velocity
Relative velocity is the speed of one object as observed from another moving object. It plays a key role in analyzing scenarios where multiple objects are moving, such as our case with the airplane and the bullet.
For the airplane to escape the bullet, its effective speed upward must be greater than the bullet's speed or such that they never meet. The concept of relative velocity can be framed as:
  • When two objects move towards each other, their relative speed is the sum of their speeds.
  • When two objects move in the same direction, their relative speed is the difference between their speeds.
In the given problem, the relative motion concept helps us understand how quickly the bullet closes in on the plane. With an upward speed of the airplane and the bullet moving in the same direction initially, it's crucial to determine if the airplane's acceleration can increase its relative velocity enough to escape being hit. To find this, we analyze the quadratic equation derived from their motion, ultimately finding the minimum acceleration needed to maintain a safe distance.

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