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.
The area between the graph and the x-axis is the distance moved. If the velocity is constant the v vs t graph is a straight horizontal line. The shape of the area under the graph is a rectangle. For constant velocity, distance = V * time. Time is the x-axis and velocity is the y-axis. If the object is accelerating, the velocity is increasing at a constant rate. The graph is a line whose slope equals the acceleration. The shape of the graph is a triangle. The area under the graph is ½ * base * height. The base is time, and the height is the velocity. If the initial velocity is 0, the average velocity is final velocity ÷ 2. Distance = average velocity * time. Distance = (final velocity ÷ 2) * time, time is on the x-axis, and velocity is on the y-axis. (final velocity ÷ 2) * time = ½ time * final velocity ...½ base * height = ½ time * final velocity Area under graph = distance moved Most velocity graphs are horizontal lines or sloping lines.
Two different distance-time graphs have matching velocity-time graphs when the slope of the distance-time graph represents the velocity in the velocity-time graph, as velocity is the derivative of distance with respect to time. This means that the steeper the distance-time graph, the greater the velocity on the velocity-time graph at that point.
The product of velocity and time yields distance travelled if the velocity is constant for the time in question. If velocity is not constant, one must first calculate the average velocity over a given time period before multiplying it by the time involved.
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.
Yes, a position-time graph can be created from a velocity-time graph by integrating the velocity values over time. By finding the area under the velocity-time curve, you can determine how the position of an object changes over time.
The area between the graph and the x-axis is the distance moved. If the velocity is constant the v vs t graph is a straight horizontal line. The shape of the area under the graph is a rectangle. For constant velocity, distance = V * time. Time is the x-axis and velocity is the y-axis. If the object is accelerating, the velocity is increasing at a constant rate. The graph is a line whose slope equals the acceleration. The shape of the graph is a triangle. The area under the graph is ½ * base * height. The base is time, and the height is the velocity. If the initial velocity is 0, the average velocity is final velocity ÷ 2. Distance = average velocity * time. Distance = (final velocity ÷ 2) * time, time is on the x-axis, and velocity is on the y-axis. (final velocity ÷ 2) * time = ½ time * final velocity ...½ base * height = ½ time * final velocity Area under graph = distance moved Most velocity graphs are horizontal lines or sloping lines.
Two different distance-time graphs have matching velocity-time graphs when the slope of the distance-time graph represents the velocity in the velocity-time graph, as velocity is the derivative of distance with respect to time. This means that the steeper the distance-time graph, the greater the velocity on the velocity-time graph at that point.
The product of velocity and time yields distance travelled if the velocity is constant for the time in question. If velocity is not constant, one must first calculate the average velocity over a given time period before multiplying it by the time involved.
In a velocity-time graph it will be the time axis (where velocity = 0). On a distance-time graph it will be a line parallel to the time axis: distance = some constant (which may be 0).
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.
To determine acceleration from a distance-time graph, calculate the slope of the graph at a specific point. The steeper the slope, the greater the acceleration. The formula for acceleration is acceleration change in velocity / time.
If an x-t graph is a position-time graph, velocity is the slope of the line on the graph.
Simply put, a velocity time graph is velocity (m/s) in the Y coordinate and time (s) in the X and a position time graph is distance (m) in the Y coordinate and time (s) in the X if you where to find the slope of a tangent on a distance time graph, it would give you the velocity whereas the slope on a velocity time graph would give you the acceleration.
A straight line on a distance - time graph represents a "constant velocity".
distance = velocity x time so on the graph velocity is slope. If slope is zero (horizontal line) there is no motion
Velocity.
Distance.