In Active Figure 2.11b, the area under the...

In Active Figure 2.11b, the area under the velocity-versus-time graph and between the vertical axis and time $t$ (vertical dashed line) represents the displacement. As shown, this area consists of a rectangle and a triangle. (a) Compute their areas. (b) Explain how the sum of the two areas compares with the expression on the right hand side of Equation 2.16 .

Key Concepts

Displacement and Velocity–Time Graphs

Displacement is obtained by integrating the velocity over a time interval, and graphically, this corresponds to the area under the velocity–time curve. This concept bridges the algebraic calculation with its geometric interpretation, making it easier to visualize and calculate displacement in problems involving motion.

Geometric Decomposition of Areas

Breaking a complicated area under a curve into simple geometric shapes such as rectangles and triangles is a valuable strategy. It enables one to compute total displacement by calculating the area of each shape separately and summing them, thereby simplifying the mathematical process even in cases of non-uniform motion.

Calculating Areas of Basic Geometric Shapes

The basic formulas for calculating areas—the area of a rectangle as base times height and the area of a triangle as one-half the product of its base and height—are essential tools in physics. They allow for the straightforward computation of displacement when the velocity–time graph is split into recognizable geometric parts.

Connection to Kinematic Equations

The sum of the computed areas from the rectangle and triangle is directly related to the kinematic formulas used in uniformly accelerated motion. This connection shows that the geometric interpretation of the area under the curve is consistent with the algebraic expression of displacement found in standard kinematic equations, confirming the validity of these relationships.

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