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 derivative
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.
Constant
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 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
Normally a position-time graph is actually a distance-time graph where the distance of an object is measured from a fixed point called the origin. The slope (gradient) of the graph is the radial velocity - or the component of the velocity in the radial direction - of the object. That is, the component of the object's velocity in the direction towards or away from the origin. Such a graph cannot be used to measure the component of the velocity at right angles to the radial direction. In particular, an object going around in a circle would appear t have no velocity since its distance from the origin remains constant.
The slope is zero.
Any graph with the slope of -1/2
The slope of the graph of [ y = x + any number ] is 1 .
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 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.
3x-5y-1 is not the equation of a line.In fact, it's not an equation at all. It's just an expression that standsfor a number. There's nothing there to represent by a graph.
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.