The remaining kinematics equations can be found by eliminating the variables v 2 and Δt. Equation 2 does not include the variable a. The width, w, is t 2 - t 1, (Δt).Īrea A 2 is a triangle with base Δt and height v 2 - v 1.Īdding areas A 2 and A 1 gives the total displacement of the object during the time interval.Įquation 1 does not include the variable s. 1, calculating the area under the line simply means calculating the area of the rectangle A 1 and the triangle A 2 and adding the values.Īrea A 1 is a rectangle. For some objects this calculation can be a little tricky, but for the object depicted in Fig. 1, which is moving at a changing velocity, can be found by calculating the area under the line during the time interval, Δt. Similarly, the total displacement of the object in Fig. The product, v 1Δt, is also equal to the area A 1. Consider a graph of the motion of this object, as in Fig. Where Δt is the time interval t 2 - t 1. The absolute value of the displacement is the distance traveled. The displacement of the object is represented by s. How can the distance be calculated for an object that is not traveling at a constant velocity?Ĭonsider an object moving with constant velocity, v 1, from time t 1 to t 2. The object in this activity, however, is not traveling at a constant velocity. The displacement for an object traveling at a constant velocity can be found by multiplying the object’s velocity by the time the object travels at that velocity. To get the next equation, derive an expression for the displacement of the object during the time interval, Δt. This expresses the equation in the slope-intercept form of a line, y = mx + b.
Rearrange Equation 1 to get v 2 on the left side of the equation. To keep things simple, rewrite t 2 - t 1 as Δt. In this case, since the slope will be a change in velocity (rise) divided by a change in time (run), the slope will equal the acceleration. The first step will be to calculate the slope of the diagonal line. Δ indicates change, for example Δv = (v 2 -v 1) S = │ s│ = d (vectors are indicated in bold the same symbol not in bold represents the magnitude of the vector) S = the displacement vector, the magnitude of the displacement is the distance, V 2 = the magnitude of the final velocity (meters per second, m/s) (in some texts this is v f)Ī = the magnitude of the acceleration (in meters per second squared, m/s 2) V 1 = the magnitude of the initial velocity (meters per second, m/s) (in some texts this is vi or v 0) V = the magnitude of the velocity of the object (meters per second, m/s) This document will make use of the following variables: The shaded area (A 1 + A 2) represents the displacement of the object during the time interval between t 1 and t 2, during which the object increased velocity from v 1 to v 2. The diagonal line represents the motion of an object, with velocity changing at a constant rate. 1, in whichthe x axis represents time and the y axis represents velocity. This exercise references the diagram in Fig. This article is a purely mathematical exercise designed to provide a quick review of how the kinematics equations are derived using algebra.
PHYSICS KINEMATICS EQUATIONS HOW TO
A solid understanding of these equations and how to employ them to solve problems is essential for success in physics. These familiar equations allow students to analyze and predict the motion of objects, and students will continue to use these equations throughout their study of physics. Kinematics is the study of the motion of objects without concern for the forces causing the motion.
0 kommentar(er)
