A derivative graph tracks the slope of a function.
The graph is a circle with a radius of 6, centered at the origin.
A line. The derivative of a function is its slope. If the slope is a constant then the graph is a line.
The steepness of a line graph is called the "gradient" ------------------------------- or slope.
velocity.
Meters/seconds squared
The slope of height vs. time squared graph equals (g) - acceleration due to gravity divided by two. In symbols m = g/2, where m is the slope and g is the acceleration due to gravity.
ds/dt gives the velocity at that instant. So slope gives the velocity
Yes. Speed is the rate at which distance changes over time. In calculus terms v = dx/dt, or the slope of the distance vs. time graph. If the slope of the distance vs. time graph is a straight line, the speed is constant.
If you have an object that is accelerating, then a position vs. time graph will give you a parabola which is pretty but is very hard to measure anything on - especially hard to measure the acceleration (or the curve of the line). If however, you graph position vs. time squared, you get a nice straight line (if you have constant acceleration) and therefore, you can measure the slope and get the acceleration. Remember: x = 1/2at2 so if you graph x vs. t2 then the slope = 1/2 a or a = 2*slope No matter what you are measuring, you always want to graph a straight line. hope that helps
That's unusual. I guess your teacher is trying to make you think a bit. It's a good mental exercise, though. You may recall that the units of acceleration are meters per second squared. That gives you a clue right there. And if you knew Calculus, you'd know that acceleration is the second derivative of distance, s, with respect to time, t: d2s/dt2. So, by now you're probably getting the feeling that the slope of a distance-time squared graph has something to do with acceleration. And you'd be right. Just as the slope of a velocity-time graph is acceleration, the slope of a distance-t2 graph is acceleration. Well, not quite. It's actually ONE HALF the acceleration.
Acceleration is the derivative of velocity (a=dv/dt). If you are not familiar with calculus then it would be sufficient to say that the slope of the line tangent to the graph, only touches at one point, is equal to the instantaneous acceleration.
No, it is a straight line passing through the origin.
The slope for a straight line graph is the ratio of the amount by which the graph goes up (the rise) for every unit that it goes to the right (the run). If the graph goes down, the slope is negative. For a curved graph, the gradient at any point is the slope of the tangent to the graph at that point.
The slope of a velocity-time graph represents acceleration.
acceleration
the slope at any point on the graph is the acceleration