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.
To calculate velocity, you need the displacement of an object (the change in position) and the time it took to make that displacement. Velocity is determined by dividing the displacement by the time taken to achieve that displacement.
Velocity is calculated by dividing the displacement of an object by the time taken to cover that displacement. It is a vector quantity, meaning it has both magnitude and direction. The formula is: velocity = displacement / time.
Your acceleration vs. Time graph is the slope of your velocity vs. time graph
To create an acceleration-time graph from a velocity-time graph, you need to find the slope of the velocity-time graph at each point. The slope represents the acceleration at that specific instant. Plot these acceleration values against time to get the acceleration-time graph.
When acceleration is constant, one equation of kinematics is: (final velocity)^2 = 2(acceleration)(displacement) + (initial velocity)^2. When you are graphing this equation with displacement or position of the x-axis and (final velocity)^2 on the y-axis, the equation becomes: y = 2(acceleration)x + (initial velocity)^2. Since acceleration is constant, and there is only one initial velocity (so initial velocity is also constant), the equation becomes: y = constant*x + constant. This looks strangely like the equation of a line: y = mx + b. Therefore, the slope of a velocity squared - distance graph is constant, or there is a straight line. Now, when you graph a velocity - distance graph, the y axis is only velocity, not velocity squared. So if: v^2 = mx + b. Then: v = sqrt(mx + b). Or: y = sqrt(mx + b). This equation is not a straight line. For example, pretend m = 1 and b = 0. So the equation simplifies to: y = sqrt(x). Now, make a table of values and graph: x | y 1 | 1 4 | 2 9 | 3 etc. When you plot these points, the result is clearly NOT a straight line. Hope this helps!
When the data on the graph is continuous,it does make sense to connect the points on the graph of 2 related variables.
Because its really fun :)
It is radial the velocity in a direction towards or away from a fixed point of reference (the origin) at a given time. The velocity time graph takes no account of motion in a direction across the radial direction.
One word answer: integrate. The area under the acceleration curve, up to time T, is the speed at time T. If you now make a curve of the speed as a function of time, and find the area under that up to time T, that will be the position at time T.
Divide the net displacement by the time of travel.
You might assume that acceleration is proportional to force - specifically, using Newton's Second Law. However, you would need to know the mass on which the force acts - otherwise, you simply don't have enough information.