Distance.
Area under velocity versus time graph(between two given instances of time i.e. two points on time axis) gives the displacement of the body( whose graph was plotted) between those two instances i.e. in that time interval. Area under velocity time graph can be found from definite integration if the graph is a curve. Note: Area under velocity versus time graph gives displacement not distance covered by body. Note: Area enclosed between the plotted curve and time axis is taken. For convenience time should be taken in the x-axis.
To determine velocity from an acceleration-time graph, you can find the area under the curve of the graph. This area represents the change in velocity over time. By calculating this area, you can determine the velocity at any given point on the graph.
The displacement of an object from a velocity-time graph can be determined by finding the area under the velocity-time graph. For example, the displacement over a certain time interval can be calculated by finding the area of the corresponding region under the velocity-time graph. This can be done by calculating the area of the trapezoid or rectangle formed by the graph.
The area under an acceleration-time graph is equal to the object's velocity (not change in velocity).
No, displacement is the area under the velocity vs. time graph. The slope of a velocity vs. time graph represents acceleration.
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 can't....it's merely impossible! Assuming it is a graph of velocity vs time, it's not impossible, it's simple. Average velocity is total distance divided by total time. The total time is the difference between finish and start times, and the distance is the area under the graph between the graph and the time axis.
Distance travelled from a velocity / time graph can be calculated from area under graph, say area under (v/t) graph from 0 - 1 seconds = distance travelled after 1 second, then do 0 - 2 seconds, 0 - 3 etc for set of data for distance / time graph
Derivitives of a velocity : time graph are acceleration and distance travelled. Acceleration = velocity change / time ( slope of the graph ) a = (v - u) / t Distance travelled = average velocity between two time values * time (area under the graph) s = ((v - u) / 2) * t
If your graph shows velocity on the vertical axis and time on the horizontal axis, then the slope of the graph represents the acceleration. More specifically, the slope of the graph at a specific point represents the acceleration at that instantaneous point in time. So if the slope of the graph doesn't change (i.e. the graph is a straight line), then the acceleration is constant and doesn't change over time. In calculus, this is represented as the derivative: The derivative of velocity with respect to time equals the acceleration.
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 area under a velocity-time graph represents the displacement of an object. If the area is positive, the object is moving in the positive direction; if negative, the object is moving in the negative direction. The steeper the slope of the graph, the greater the velocity.