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Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object. (Velocity is on the y-axis and time on the x-axis. Multiplying the velocity by the time, the time cancels out, and only displacement remains.)
These relationships can be demonstrated graphically. The gradient of a line on a displacement time graph represents the velocity. The gradient of the velocity time graph gives the acceleration while the area under the velocity time graph gives the displacement. The area under a graph of acceleration versus time is equal to the change in velocity.
From this derivative equation, in the one-dimensional case it can be seen that the area under a velocity vs. time (v vs. t graph) is the displacement, s. In calculus terms, the integral of the velocity function v(t) is the displacement function s(t). In the figure, this corresponds to the yellow area under the curve.
Trajectory of a particle with initial position vector r 0 and velocity v 0, subject to constant acceleration a, all three quantities in any direction, and the position r(t) and velocity v(t) after time t. The initial position, initial velocity, and acceleration vectors need not be collinear, and the equations of motion take an almost identical ...
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The areal velocity magnitude (i.e., the areal speed) is this region's area divided by the time interval Δt in the limit that Δt becomes vanishingly small. The vector direction is postulated to be normal to the plane containing the position and velocity vectors of the particle, following a convention known as the right hand rule.
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Many other fundamental quantities in science are time derivatives of one another: force is the time derivative of momentum; power is the time derivative of energy; electric current is the time derivative of electric charge; and so on. A common occurrence in physics is the time derivative of a vector, such as velocity or displacement. In dealing ...