Want this question answered?
The area under the velocity time graph of an object is equal to the distance travelled by that object in that time. This is because displacement is the integral of velocity with respect to time so integrating velocity from time A to time B will give the displacement from time A to time B. ( Integrating is the same as calculating the area under the 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.
It is the force constant of the material in N/m. So you can substitute it into the equation F=kx (F=force, k=force constant or gradient in N/m, x = extension) You would expect the extension to be on the y-axis normally since it is the measured value. However since you want to use the graph to calculate certain values it is on the x-axis (you can also find the work done by the force by finding the area under the graph) Also it allows you to divide the y-axis values by the cross-sectional area and x-axis values by original length to get a stress vs strain graph where you can use the gradient to find the Young modulus of the material.
Mechanics is a branch of science that dwells on the behavior of physical bodies and how they behave when under force or displacement.
False
Work done by the force.
It is the work done or the energy utilised
Displacement is the area under the v-t graph.
The displacement, along the direction of measurement, is zero. It need not mean that the object is back at the starting point. The displacement-time graph, measuring the vertical displacement of a ball thrown at an angle, will have displacement = 0 when the ball returns to ground level but, unless you are extremely feeble, the ball will be some distance away, not at its starting point which is where you are. The use of such a graph is not unusual in the elementary projectile motion under gravity.
The momentum-time graph is the integral of the force-time graph. that is, it is the area under the curve of the f-t graph.The momentum-time graph is the integral of the force-time graph. that is, it is the area under the curve of the f-t graph.The momentum-time graph is the integral of the force-time graph. that is, it is the area under the curve of the f-t graph.The momentum-time graph is the integral of the force-time graph. that is, it is the area under the curve of the f-t graph.
The area under a graph of force against distance (or extension, if it's a spring) represents the work done by that force. Since it sounds like you're talking about a spring, you should know that the area would represent the work done to stretch the spring that distance, and also represents the amount of elastic potential energy contained by the spring.
It is not, if it is a graph of force against acceleration.
no, work done is the area under a force-distance graph
WORK
The area of a position-time graph does not have a meaning. However, the area under a velocity-time graph is the displacement. Refer to the related link below for an illustration.
The area under the velocity time graph of an object is equal to the distance travelled by that object in that time. This is because displacement is the integral of velocity with respect to time so integrating velocity from time A to time B will give the displacement from time A to time B. ( Integrating is the same as calculating the area under the graph)
You cannot since the graph shows displacement in the radial direction against time. Information on transverse displacement, and therefore transverse velocity, is not shown. For example, there is no difference in the graph of you're staying still and that of your running around in a circle whose centre is the origin of the graph. In both cases, your displacement from the origin does not change and so the graph is a horizontal line. In the first case the velocity is 0 and in the second it is a constantly changing vector. All that you can find is the component of the velocity in the radial direction and this is the slope of the graph at the point in question.