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 from
a point x1, to point x2, represents the work done by the force;
W =
⌠Fdx
If the force is constant from x1 to x2, then; W =
F∙(x
2 - x1)
The slope of the curve at a given value of x, (dF
/dx),
tells us how the force F is
varying 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.
If you are applying a constant net force to an object and measuring the acceleration as you change mass, the graph of acceleration (y-axis, dependent variable) vs mass (x-axis, independent variable) would be hyperbolic. Acceleration is inversely proportional to mass. If you then graph acceleration (on the y-axis) vs 1/mass (on the x-axis), the slope would be your constant net force (Fnet). For a graph of 1/mass (on the y-axis) vs acceleration (on the x-axis), the slope would be 1/Fnet. This relationship is shown by Newton's second law, Fnet=ma or a=Fnet/m.
Nothing. Because Force divided by mass is acceleration and you get expect to get something by taking gradient of acceleration-acceleration graph. Most probably you mistyped the question. Check again so that a better answer may be provided.
Acceleration vs. mass is a curved graph. You need calculus to figure that out and it depends at what point on the graph you want the slope. You have to use the process of differentiating.
^ Wow dude... No WAY! -_-
You don't need differentiating at all. The gradient of the graph represents the force acting on the mass. First you need to make the graph linear, by making it a- 1/mass (on the X axis) vs acceleration ms-2 (on the Y axis). Then you have a linear graph. By using the Calculus formula Y2-Y1/X2-X1 you can get the gradient of the graph, which represents the force acting on the mass in N (newtons)
It is equal to the mass of the object being accelerated.
f = ma y = mx (m = gradient)
da frenchman gogo it represents the meanings of the third force called muth. with counteract the action of both of these amazing amzing forces. they make me giidy.
I believe that would be force.
Yes it does. Velocity = Displacement / Time. On a graph of displacement vs time, the slope is the velocity. Steeper slope = higher velocity, flatter slope = lower velocity.
False. The slope of a velocity vs time graph is acceleration
The slope of that graph at each point is the speed at that instant of time.
False
The slope of a velocity-time graph represents acceleration.
The slope at each point of a displacement/time graph is the speed at that instant of time. (Not velocity.)
Yes it does. Velocity = Displacement / Time. On a graph of displacement vs time, the slope is the velocity. Steeper slope = higher velocity, flatter slope = lower velocity.
It is the instantaneous speed in the direction in which the displacement is measured.
False. The slope of a velocity vs time graph is acceleration
The slope of a velocity-time graph represents acceleration.
The slope of that graph at each point is the speed at that instant of time.
False
The slope of a velocity-time graph represents acceleration.
The slope of a velocity-time graph represents acceleration.
The slope of a distance-time graph represents speed.
A displacement vs. time graph of a body moving with uniform (constant) velocity will always be a line of which the slope will be the value of velocity. This is true because velocity is the derivative (or slope at any time t) of the displacement graph, and if the slope is always constant, then the displacement will change at a constant rate.
The slope of the function on a displacement vs. time graph is (change in displacement) divided by (change in time) which is just the definition of speed. A relatively steep slope indicates a relatively high speed.