Calculating Tangent for Accelerated Motion: Rise/Run?

[Turvey]

Just making sure that the proper forumla to calculate a tangent on a graph, (Position vs. Time) that shows an accelerated motion is rise/run.

And also, I would like to make sure that if it asks you to start at the point 0.15s, that as long as you draw the line to cross a point without going through the line, you can virtually pick any point on the x and y-axis to produce the correct answer, as long as the line touches the starting point on the x axis.

Thanks for reading, and any response is greatly appreciated.

 

[HallsofIvy]

Just making sure that the proper forumla to calculate a tangent on a graph, (Position vs. Time) that shows an accelerated motion is rise/run.

And also, I would like to make sure that if it asks you to start at the point 0.15s, that as long as you draw the line to cross a point without going through the line, you can virtually pick any point on the x and y-axis to produce the correct answer, as long as the line touches the starting point on the x axis.

Thanks for reading, and any response is greatly appreciated.

This not at all clear. You talk about "graphs" but you seem to be thinking specifically about motion. If you have a "distance versus time" graph, with constant acceleration, then the graph is NOT a straight line, the slope of the tangent line at any point varies with the point, and the slope is not "rise over run" since neither is those is defined at a single point. The slope of the tangent line can be calculated by taking the derivative.

It is true that for a straight line, you can use whatever (x,y) points you want (on the line of course) to calculate the slope- that's true since the slope of a straight line IS a single number and is independent of the points you use to calculate (y₁- y₀)/(x₁-x₀). SO if you (accurately) draw the tangent line (that may be what you meant by "draw the line to cross a point without going through the line") then you can use any two points on that tangent line to find the slope of the tangent line.

(Though I might point out that the "tangent line" to y= x³ at x= 0 does "go through" the graph!)