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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.)
There are two main descriptions of motion: dynamics and kinematics.Dynamics is general, since the momenta, forces and energy of the particles are taken into account. In this instance, sometimes the term dynamics refers to the differential equations that the system satisfies (e.g., Newton's second law or Euler–Lagrange equations), and sometimes to the solutions to those equations.
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.
Consequently, the acceleration is the second derivative of position, [7] often written . Position, when thought of as a displacement from an origin point, is a vector: a quantity with both magnitude and direction. [9]: 1 Velocity and acceleration are vector quantities as well. The mathematical tools of vector algebra provide the means to ...
Snap, [6] or jounce, [2] is the fourth derivative of the position vector with respect to time, or the rate of change of the jerk with respect to time. [4] Equivalently, it is the second derivative of acceleration or the third derivative of velocity, and is defined by any of the following equivalent expressions: = ȷ = = =.
In terms of a displacement-time (x vs. t) graph, the instantaneous velocity (or, simply, velocity) can be thought of as the slope of the tangent line to the curve at any point, and the average velocity as the slope of the secant line between two points with t coordinates equal to the boundaries of the time period for the average velocity.
By the fundamental theorem of calculus, it can be seen that the integral of the acceleration function a(t) is the velocity function v(t); that is, the area under the curve of an acceleration vs. time (a vs. t) graph corresponds to the change of velocity. =.
For rod length 6" and crank radius 2" (as shown in the example graph below), numerically solving the acceleration zero-crossings finds the velocity maxima/minima to be at crank angles of ±73.17615°. Then, using the triangle law of sines, it is found that the rod-vertical angle is 18.60647° and the crank-rod angle is 88.21738°. Clearly, in ...
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