Kinematics, the study of motion, provides a framework for describing how objects move. Calculating the velocity of a moving object is a fundamental task in physics, often requiring knowledge of the time an object spent accelerating or traveling. A specific challenge arises when a scenario provides information about an object’s change in position and acceleration, but not the duration of the motion.
The Third Kinematic Equation
Physics uses a specialized tool, often referred to as the third kinematic equation, to solve for velocity when the time variable is missing. This relationship mathematically links an object’s final velocity, initial velocity, acceleration, and displacement. The equation is represented as: \(v_f^2 = v_i^2 + 2a\Delta x\). This form bypasses the need for the time variable (\(t\)), ensuring that one can analyze constant acceleration motion even when the duration is unknown.
Defining the Variables
To utilize this equation effectively, one must first understand the meaning and standard units of each variable. The final velocity (\(v_f\)) is the speed and direction of the object at the end of the observed motion, and the initial velocity (\(v_i\)) is the speed and direction at the starting point. Both are vector quantities, measured in meters per second (m/s), and their direction must be considered, usually represented by a positive or negative sign.
The acceleration (\(a\)) is the rate at which velocity changes, measured in meters per second squared (\(m/s^2\)). A positive value means the object is speeding up in the positive direction, while a negative value indicates slowing down or speeding up in the negative direction. Displacement (\(\Delta x\)) is the overall change in the object’s position, measured in meters (m). Displacement is a vector representing the straight-line distance from the start point to the end point.
Step-by-Step Calculation
The first step is to clearly identify the known variables: initial velocity, acceleration, or displacement. The equation \(v_f^2 = v_i^2 + 2a\Delta x\) must then be algebraically rearranged to isolate the unknown velocity. If finding the final velocity, the equation is ready; otherwise, rearrangement is necessary, such as \(v_i^2 = v_f^2 – 2a\Delta x\) for initial velocity.
Next, substitute the known numerical values into the expression. It is crucial to ensure that all units are consistent, typically using SI units of meters and seconds, before performing the arithmetic. After substituting the numbers, the algebraic operations are carried out to find the squared velocity.
The final step is taking the square root of the result, which yields the numerical magnitude and direction of the unknown velocity. The square root introduces both a positive and a negative value. This dual solution is physically significant because velocity is a vector, and the two signs represent the two possible directions the object could be moving.
Worked Example Application
Consider a car braking over a displacement of \(30.0 \ m\) with a constant acceleration of \(-5.0 \ m/s^2\) (negative because it is slowing down). If the final velocity (\(v_f\)) is \(10.0 \ m/s\), we aim to find the initial velocity (\(v_i\)). The known values are \(v_f = 10.0 \ m/s\), \(a = -5.0 \ m/s^2\), and \(\Delta x = 30.0 \ m\).
The equation is rearranged to solve for \(v_i^2\): \(v_i^2 = v_f^2 – 2a\Delta x\). Substituting the known values yields: \(v_i^2 = (10.0 \ m/s)^2 – 2(-5.0 \ m/s^2)(30.0 \ m)\).
The calculation proceeds as follows: \(v_i^2 = 100 \ (m/s)^2 – (-300 \ (m/s)^2)\), which simplifies to \(v_i^2 = 400 \ (m/s)^2\). Taking the square root provides the initial velocity magnitude as \(20.0 \ m/s\). Since the car was slowing down while moving forward, the positive value is the physically meaningful result, indicating the initial speed was \(20.0 \ m/s\).