/*! 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 74 The position of a particle at an... [FREE SOLUTION] | 91Ó°ÊÓ

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The position of a particle at any instant \(t\) is given by \(x=t^{2}-7 t\). The average velocity of the partielc from \(t=t_{1}\) to \(t=t_{2}\) is zero. Here \(t_{1}\), and \(t_{2}\) respectively, cannot be cqual to: (a) 1,6 (b) \(2.5\) (c) \(2.5,3.5\) (d) 0,7

Short Answer

Expert verified
(c) 2.5, 3.5

Step by step solution

01

Understanding the problem

We are asked to find the values of \(t_1\) and \(t_2\) for which the average velocity of a particle is zero. The position of the particle is given by \(x=t^2 - 7t\). Average velocity over the interval \([t_1, t_2]\) is given by \(\frac{x(t_2) - x(t_1)}{t_2 - t_1} = 0\). This implies \(x(t_2) = x(t_1)\).
02

Finding position equation

Given the position equation \(x = t^2 - 7t\), we evaluate it for \(t_1\) and \(t_2\) to find \(x(t_1) = t_1^2 - 7t_1\) and \(x(t_2) = t_2^2 - 7t_2\).
03

Setting the equation to zero

Since average velocity is zero, \(x(t_2) - x(t_1) = 0\). Therefore, \(t_2^2 - 7t_2 = t_1^2 - 7t_1\). Simplifying, we get \(t_2^2 - t_1^2 = 7(t_2 - t_1)\).
04

Factoring the expression

Using the identity \(a^2 - b^2 = (a-b)(a+b)\), we rewrite the equation as \((t_2-t_1)(t_2+t_1) = 7(t_2 - t_1)\). If \(t_2 eq t_1\), we can divide both sides by \((t_2-t_1)\) yielding \(t_2 + t_1 = 7\).
05

Finding incompatible values

The equation \(t_2 + t_1 = 7\) defines a condition for zero average velocity. For options (a) through (d):- (a) \(1 + 6 = 7\), so the velocity condition holds.- (b) involves only one value, so cannot be zero velocity.- (c) \(2.5 + 3.5 = 6\), does not satisfy \(t_2 + t_1 = 7\).- (d) \(0 + 7 = 7\), so the condition holds. Hence, (c) is the correct answer.

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Key Concepts

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

Position Equation
The position of a particle at any time, or instant, can be described using a position equation. In this exercise, the position equation is given by \( x = t^2 - 7t \). This equation relates the position of the particle \( x \) to time \( t \). The term \( t^2 \) accounts for changes that grow quadratically with time, while \( -7t \) accounts for linear changes over time. Altogether, this equation allows us to precisely determine where the particle is at any given moment. To better understand the behavior of the particle, we can substitute specific values of \( t \) into the equation. This helps us see how "position" varies over time.
Factoring Expressions
Factoring expressions is a technique used to simplify algebraic expressions or equations. It involves rewriting an equation as a product of simpler expressions. In our problem, we have an expression involving two squares: \( t_2^2 - t_1^2 \). We can rewrite this by using the factoring method known as the difference of squares. This method expresses the difference between two squares as a product of two binomials. Specifically, the identity \( a^2 - b^2 = (a-b)(a+b) \) is used. Applying this to our expression gives us: \( (t_2 - t_1)(t_2 + t_1) \). Factoring helps us identify important relationships or simplify calculations in many algebra problems.
Identity for Differences of Squares
The identity for differences of squares is a powerful tool in mathematics, particularly when simplifying expressions or solving equations. This identity states that any difference of squares can be factored as \( a^2 - b^2 = (a-b)(a+b) \). It's important because it transforms a subtraction into a multiplication, making it a useful simplification technique. In the given problem, the expression \( t_2^2 - t_1^2 \) fits this pattern perfectly. Thus, you can rewrite it as \( (t_2-t_1)(t_2+t_1) \). Applying this identity allows us to manipulate and solve equations more easily, especially when solving for points where functions equal zero or determining certain values of variables.
Zero Average Velocity Condition
The zero average velocity condition is key to solving the exercise. It occurs when the position of a particle at two different time points results in no net movement over the interval. Mathematically, this is expressed as the change in position equalling zero, or \( x(t_2) = x(t_1) \).
  • For average velocity to be zero, the equation simplifies to the condition \( t_2 + t_1 = 7 \) once other terms are canceled or factored out.
  • This equation indicates that any pair of \( t_1 \) and \( t_2 \) with this sum will result in zero average velocity.
This can be applied to identify viable time pairs in options like (a), which hold this property, while ensuring pairs like in option (c) (where the sum is not 7) do not. Understanding and applying this condition helps determine when the particle returns to a previous position or when there is no net change in position over a time interval.

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