1/Time?
At any given point the relationship would be (D/t) / D where D is distance and t is time.
D/t / D = (D/t)(1/D) = D / Dt = 1/t.
If however you plot position v. time then the slope is speed (not velocity, because the graph does not specify direction).
The tangent at a point on the position-time graph represents the instantaneous velocity. 1. The tangent is the instantaneous slope. 2. Rather than "average" velocity, the slope gives you "instantaneous" velocity. The average of the instantaneous gives you average velocity.
The slope of the line of a speed-versus-time graph will give you acceleration. Remember that acceleration may be positive or negative, and in some cases, acceleration may be positive when speed remains the same.1 If the speed-time curve is linear or piecewise linear2, acceleration is, as stated above, merely the slope of the line segment. If, however, the graph is a smooth curve, then changing acceleration is represented. In other words, the rate of change of velocity -- delta-V over delta-T -- is not a constant. In that case, the slope of the line segment tangent to the curve at any given point is the acceleration at that point. Note 1: There is a discussion comment on this point.Note 2: See the web link for an example of a graph that is piecewise linear.
If the slope of a line on a distance-time graph is 1, it means that the speed of the object being plotted is 1 unit of distance traveled per unit of time elapsed. So, if the units are in, for example, meters and seconds, the speed would be 1 meter per second.
if the segments on the disp vs time graph are straight lines, you merely measure the slope of those lines; the velocity is the slope of the lineso if the disp vs time graph shows a straight line of slope 3 between say t=0 and t=4, then you know the object had a constant speed of 3 units between t=0 and t=4;if the disp vs time graph is curved, then you need to find the slope of the tangent line to the disp vs time curve at each point; the slope of this tangent line is the instantaneous speed at the time, and with several such measurements you can construct your v vs t graph
A curve of a force F, vs displacement x (F vs x), represents the magnitude of a force as it is producing a displacement of a body. The area under the curve froma point x1, to point x2, represents the work done by the force;W =⌠FdxIf the force is constant from x1 to x2, then; W =F∙(x2 - x1)The slope of the curve at a given value of x, (dF/dx),tells us how the force F isvarying with displacement x at that point.For the case of a constant force, the value of the slope is zero, (dF/dx=0),meaning that the force is not varying as the displacement takes place.
The tangent at a point on the position-time graph represents the instantaneous velocity. 1. The tangent is the instantaneous slope. 2. Rather than "average" velocity, the slope gives you "instantaneous" velocity. The average of the instantaneous gives you average velocity.
The slope is zero.
Any graph with the slope of -1/2
To determine the order of reaction from a graph, you can look at the slope of the graph. If the graph is linear and the slope is 1, the reaction is first order. If the slope is 2, the reaction is second order. If the slope is 0, the reaction is zero order.
The slope of the graph of [ y = x + any number ] is 1 .
The equation has no slope. The graph of the equation is a straight line with a slope of -1 .
Since distance is 1/2 at^2 where a is acceleration, it represents one half of the acceleration
1
5
It is 6 and the y intercept is -1
The slope of the graph of that equation is -1.
1) You write the equation in slope-intercept form, if it isn't in that form already. 2) An easy way to graph it is to start with the y-intercept. For example, if the intercept is +5, you graph the point (0, 5). Then you add an additional point, according to the slope. For example, if the slope is 1/2, you go 2 units to the right, and one up, and graph a point there.