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The area under a v/t graph is how far you've gone. Choose a point on the time axis, read off the speed and find the area underneath. If its a straight line graph, all you have to do is find the area of the triangle. This area is the distance travelled in this particular time. Repeat for several more points on the time axis. Plot distance against time.
What else can I help you with?
How do you go from a position graph to a velocity 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.
How do you make an acceleration time graph from the data of a velocity 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.
What do slopes of straight lines on a position vs time graph mean?
The slope of a straight line on a position vs. time graph represents the object's velocity. A steeper slope indicates a faster velocity, while a shallower slope indicates a slower velocity. The slope can be positive for motion in the positive direction and negative for motion in the negative direction.
How can a distance be calculated from the v-t graph?
To calculate the distance from a velocity-time graph, you can find the area under the graph. If the graph forms a triangle, you can use the formula for finding the area of a triangle (0.5 * base * height). If the graph forms other shapes, you can break down the area into smaller, more manageable shapes and calculate each separately before summing them up.
How to calculate distance from a velocity time 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.
What is a graph with time plotted on the x axis and distance plotted on the y axis called?
It could be a velocity graph or an acceleration graph. If the plot is a straight line it is constant velocity. If the plot is a curve it is acceleration.
How do you go from a position graph to a velocity 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.
How do you make an acceleration time graph from the data of a velocity 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.
Difference between velocity time and distance time?
Velocity is distance divided by time. So the value of the velocity-time plot at any point in time will be the slope of the distance-time plot at that point in time.
Suppose you plot the distance traveled by an object at various times and you discover that the graph is not a straight line what does this indicate about the object acceleration?
If the graph of distance traveled vs. time is not a straight line, it indicates that the object's acceleration is not constant. Acceleration is the rate of change of velocity, so a non-linear distance-time graph suggests that the object's velocity is changing at a non-constant rate, causing a curved graph.
How do you find final position in an velocity vs time graph?
I think you mean distance traveled. Every tiny period "dt" of time, the distance gone is the velocity at that time, times dt. Plot velocity against time. Each little slice of velocity times dt is a slice of the area. So the total distance is the total area under the graph from time t=0 to the finish, or to whatever time you want. This is the principle behind the Integral Calculus.
How do you generate a position-time graph from a velocity time graph?
A girl walks along a straight path to drop a letter in the letterbox and comes back to his initial position. Her displacement-time graph. Plot a velocity-time graph for the same
How do you measure constant velocity?
Constant velocity is, well, constant. To measure it, measure the displacement through a given or fixed period of time. You'll have distance and time. Distance per unit time is speed. Distance per unit time (speed) with a direction vector is velocity. Velocity is speed in a given direction. If something is moving at constant velocity, it is moving at a constant speed in one direction. No changes in speed (no positive or negative acceleration, or, said another way, no acceleration at all), and no change in direction or heading.
What is a distance vs time graph?
A distance vs time squared graph shows shows the relationship between distance and time during an acceleration. An example of an acceleration value would be 3.4 m/s^2. The time is always squared in acceleration therefore the graph can show the rate of which an object is moving
How does your distance time graph look alike?
It is a zig-zag plot.
How do you find distance given changing velocity time and uniform acceleration?
Two ways. Use an equation of motion: u=initial velocity, v=final velocity, a=acceleration, t= time and s=displacement: v=u+at s=ut+1/2at2 v2- u2 = 2as Or, plot a v/t graph and find the area underneath it at a particular time.
Why is the slope of a distance velocity squared graph straight and a distance velocity graph is not?
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!
