Question

When an object is released from rest, the distance fallen between time \(t_{1}\) and time \(t_{2}\) is \(\frac{1}{2} g t_{2}^{2}-\frac{1}{2} g t_{1}^{2},\) where \(g\) is the acceleration due to gravity. Factor this expression.

Short answer

\(\frac{1}{2} g (t_{2}+t_{1})(t_{2}-t_{1})\)

Step-by-step solution

Understand the given expression

Recognize that the given expression represents the distance fallen by an object in the time interval between \(t_1\) and \(t_2\) under the influence of gravity, with \(g\) being the acceleration due to gravity. The expression is \(\frac{1}{2} g t_{2}^{2}-\frac{1}{2} g t_{1}^{2}\).

Factor the common terms

Identify and factor out the common terms from each part of the expression. Both terms contain \(\frac{1}{2} g\), which can be factored out. This leaves us with the expression \(\frac{1}{2} g (t_{2}^{2}-t_{1}^{2})\).

Apply the difference of squares

Recognize that \(t_{2}^{2}-t_{1}^{2}\) is a difference of squares, which can be factored into \((t_{2}+t_{1})(t_{2}-t_{1})\). The fully factored form of the expression is therefore \(\frac{1}{2} g (t_{2}+t_{1})(t_{2}-t_{1})\).

Key concepts

Acceleration Due to Gravity

Understanding the concept of acceleration due to gravity is key for various physics calculations related to motion. Gravity is a natural phenomenon by which all things with mass or energy are brought toward one another, and on Earth, it gives weight to physical objects.

Acceleration due to gravity, denoted by the symbol 'g', is the acceleration that the Earth imparts to objects on or near its surface. Its average value on the surface of the Earth is approximately \( 9.81 \text{ m/s}^2 \). However, this value can vary slightly depending on altitude and geographical location.

When an object is dropped, it accelerates downward, and 'g' represents the rate at which this acceleration increases. The object's velocity increases in the direction of gravity—downwards—by \(9.81 \text{ m/s}^2 \) every second, assuming air resistance is negligible. This acceleration is constant, meaning it doesn't change as the object falls, which simplifies the equations of motion for falling objects.

Distance Fallen by an Object

The distance an object falls is an essential concept in kinematics, the branch of physics that describes the motion of objects. The equation used in our exercise, \( \frac{1}{2} g t_{2}^{2}-\frac{1}{2} g t_{1}^{2} \), actually comes from the kinematic formulas that describe motion with a constant acceleration — in this case, gravity.

To understand the distance fallen, one must recognize that it's the change in position of the object. When released from rest, the initial velocity is zero, and the only force acting on the object is gravity. The distance the object falls during a specific time period can be calculated using the formula for uniformly accelerated motion: \( s = \frac{1}{2} g t^{2} \), where 's' is the distance, 'g' is the acceleration due to gravity, and 't' is the time during which the object falls.

In our problem, the total distance fallen from rest after time \( t_2 \) is \( \frac{1}{2} g t_{2}^{2} \), and the distance after time \( t_1 \) is \( \frac{1}{2} g t_{1}^{2} \). The expression provided denotes the distance fallen between the two times, \( t_1 \) and \( t_2 \). Important note: the motion is assumed to be free of air resistance and other external forces.

Difference of Squares

The difference of squares is a key mathematical concept employed in the factoring of algebraic expressions. It refers to the form \( a^2 - b^2 \), which can be factored into \( (a + b)(a - b) \).

This is an essential technique in simplifying the expressions that include squared terms subtracted from one another, as seen in the exercise provided.

Breaking Down Difference of Squares

To apply this to our physics problem, we look at the terms \( t_{2}^{2} \) and \( t_{1}^{2} \). According to the difference of squares formula, this can be rewritten as \( (t_2 + t_1)(t_2 - t_1) \). The factored form can be extremely useful in solving equations or simplifying expressions within physics calculations.

Recognizing such patterns allows for the simplification of equations, which then makes it easier to understand the physical phenomena being described — in this case, the motion of a freely falling object under the influence of Earth's gravity.