You can calculate the tangent for a give time, T, as follows:
Substitute the value of the time in the distance-time equation to find the distance at the given time. Suppose it is f(T).
Differentiate the distance-time equation with respect to time. For any given time, substitute its value in the derivative and evaluate. That is the gradient of the tangent, v.
Then equation of the tangent is
f(T) - f(t) = v*(T - t)
A tangent to a velocity-time graph represents the instantaneous acceleration of an object at that specific moment in time. It shows how the velocity is changing at that particular point.
To calculate distance from a velocity-time graph, you would find the area under the curve, as this represents the displacement or distance traveled. If the graph is above the time axis, calculate the area above the time axis, and if it dips below, calculate the area below the time axis. Summing these two areas will give you the total distance traveled.
To find instantaneous velocity from a position-time graph, you calculate the slope of the tangent line at a specific point on the graph. The slope represents the rate of change of position at that instant, which is equivalent to the velocity at that particular moment.
The slope of a distance versus time graph represents the speed or velocity of an object. A steeper slope indicates a higher speed, while a gentler slope indicates a slower speed. If the slope is negative, it means the object is moving in the opposite direction.
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.
Slope = change in y (distance) / change in x (time). If the graph is not a straight line then either apply the above formula to the tangent at the point of interest or differentiate the equation of the graph.
The variable plotted along the vertical axis is the distance in the first case, speed in the second. The gradient of (the tangent to) the distance-time graph is the speed while the area under the curve of the speed-time graph is the distance.
speed is the gradient under the distance vs time graph which is change in distance /change in time
That's not correct. If you have a graph of distance as a function of time, the speed is the slope of the graph.
The gradient (slope) of the tangent to the graph at the given time - provided that it exists. If the graph is a straight line at that point, it is the gradient of that line.
To get speed from a distance-time graph, you would calculate the slope of the graph at a given point, as the gradient represents speed. To calculate total distance covered, you would find the total area under the graph, as this represents the total distance traveled over time.
Speed (in the radial direction) = slope of the graph.
Probably: Average Speed = Total Distance/Total Time. or Instantaneous Speed = Gradient of the tangent to the Distance v Time graph.
The instantaneous speed at a specific point on a speed-time graph is the slope of the tangent to the curve at that point. It represents the speed of an object at that exact moment in time. This can be determined by calculating the gradient at that particular point.
If you graph distance vs. time, the slope of the line will be the average speed.
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.
instantaneous acceleration* * * * *No it does not.The graph is a distance-time graph so the coordinates of a point on the graph represent the position (distance) at the specified time. The gradient of the tangent to the curve at that point represents the instantaneous radial velocity. The second derivative at that point, if it exists, would represent the acceleration.