Question

why is the graph of the force vs position not uniform

Answer 1

Given below is a strobe picture of a ball rolling across a table. Strobe pictures reveal the position of the object at regular intervals of time, in this case, once each 0.1 seconds. Notice that the ball covers an equal distance between flashes. Let's assume this distance equals 20 cm and display the ball's behavior on a graph plotting its x-position versus time. The slope of the position versus time graph shown above would equal 20 cm divided by 0.1 sec or 200 cm/sec. The following graph displays this exact same information in a new format, a velocity versus time graph. This graph very clearly communicates that the ball's velocity never changes since the slope of the line equals zero. Note that during the interval of time being graphed, the ball maintained a constant velocity of 200 cm/sec. We can also infer that it is moving in a positive direction since the graph is in quadrant I where velocities are positive. To determine how far the ball travels on this type of graph we must calculate the area bounded by the "curve" and the x- or time axis. As you can see, the area between 0.1 and 0.3 seconds confirms that the ball experienced a displacement of 40 cm while moving in a positive direction. Given below are three orientations of velocity-time graphs for one-dimensional uniform velocity. On each graph, the height of the graph represents the object's velocity and the area bounded by the graph and the x- or time axis represents the object's displacement, or change in position. v vs t - since its slope equals zero there is no acceleration, or change in velocity. The object is traveling at a constant, steady rate. It is moving in a positive direction since the graph is in quadrant I where the y-axis (aka, velocity value) is positive. We know the object was traveling in a positive direction since its rectangular area is located in a positive quadrant. v vs t - since its slope equals zero there is no acceleration, or change in velocity. This object is NOT moving since its velocity equals zero. The object is in a state of rest and obviously has no displacement. v vs t - since its slope equals zero there is no acceleration, or change in velocity. The object is traveling at a constant, steady rate. It is moving in a negative direction since the graph is in quadrant IV where the y-axis (aka, velocity value) is negative. We know the object was traveling in a negative direction since its rectangular area is located in a negative quadrant.