how to find position from velocity graph

Determining the Area on a v-t Graph

As learned in an earlier part of this lesson, a plot of velocity-time can exist used to determine the dispatch of an object (the slope). In this office of the lesson, nosotros will learn how a plot of velocity versus fourth dimension can likewise exist used to decide the displacement of an object. For velocity versus time graphs, the area spring by the line and the axes represents the displacement. The diagram beneath shows three unlike velocity-time graphs; the shaded regions between the line and the time-axis represent the displacement during the stated fourth dimension interval.

The method used to observe the area under a line on a velocity-time graph depends upon whether the section jump by the line and the axes is a rectangle, a triangle or a trapezoid. Expanse formulas for each shape are given below.

Rectangle	Triangle	Trapezoid
Area = b • h	Expanse = ½ • b • h	Area = ½ • b • (h_i + h₂)

Calculating the Expanse of a Rectangle

Now we will look at a few example computations of the area for each of the to a higher place geometric shapes. First consider the calculation of the expanse for a few rectangles. The solution for finding the surface area is shown for the showtime example beneath. The shaded rectangle on the velocity-time graph has a base of half dozen s and a height of thirty yard/s. Since the expanse of a rectangle is found by using the formula A = b 10 h, the area is 180 g (6 due south ten xxx thousand/southward). That is, the object was displaced 180 meters during the first 6 seconds of motion.

Area = b * h
Area = (vi s) * (30 m/south)

Surface area = 180 grand

At present try the post-obit two practice problems every bit a check of your understanding. Make up one's mind the displacement (i.e., the area) of the object during the beginning 4 seconds (Practice A) and from three to vi seconds (Exercise B).

Calculating the Area of a Triangle

Now we will look at a few instance computations of the area for a few triangles. The solution for finding the area is shown for the starting time case beneath. The shaded triangle on the velocity-time graph has a base of four seconds and a superlative of forty m/southward. Since the area of triangle is found past using the formula A = ½ * b * h, the area is ½ * (four s) * (40 m/s) = 80 m. That is, the object was displaced fourscore meters during the four seconds of motion.

Area = ½ * b * h
Area = ½ * (iv s) * (40 thou/due south)

Area = eighty m

Now effort the following two practice bug as a check of your understanding. Determine the deportation of the object during the get-go 2nd (Practice A) and during the start 3 seconds (Practise B).

Computing the Expanse of a Trapezoid

Finally we will await at a few instance computations of the area for a few trapezoids. The solution for finding the expanse is shown for the commencement example below. The shaded trapezoid on the velocity-time graph has a base of 2 seconds and heights of x thou/s (on the left side) and xxx chiliad/due south (on the right side). Since the area of trapezoid is found by using the formula A = ½ * (b) * (h₁ + h_ii), the expanse is 40 m [½ * (two s) * (10 m/due south + 30 m/south)]. That is, the object was displaced 40 meters during the time interval from 1 second to 3 seconds.

Surface area = ½ * b * (h_ane + h₂)
Area = ½ * (2 s) * (10 m/s + 30 one thousand/s)

Area = twoscore m

Now try the following 2 practice problems as a check of your understanding. Determine the deportation of the object during the time interval from 2 to 3 seconds (Do A) and during the first 2 seconds (Practise B).

Culling Method for Trapezoids

An alternative ways of determining the area of a trapezoid involves breaking the trapezoid into a triangle and a rectangle. The areas of the triangle and rectangle tin be computed individually; the area of the trapezoid is so the sum of the areas of the triangle and the rectangle. This method is illustrated in the graphic below.

Triangle: Surface area = ½ * (2 due south) * (20 m/south) = 20 grand

Rectangle: Area = (2 s) * (ten m/due south) = 20 chiliad

Total Expanse = 20 g + 20 m = 40 k

It has been learned in this lesson that the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that item time period. The area can be identified as a rectangle, triangle, or trapezoid. The area can be subsequently adamant using the appropriate formula. Once calculated, this area represents the displacement of the object.

Investigate!

The widget below computes the surface area between the line on a velocity-time plot and the axes of the plot. This expanse is the displacement of the object. Apply the widget to explore or only to practice a few self-made problems.

We Would Like to Suggest ...

Sometimes it isn't plenty to just read about information technology. You have to collaborate with it! And that'due south exactly what you do when you use ane of The Physics Classroom's Interactives. We would like to propose that yous combine the reading of this page with the use of our Two Stage Rocket Interactive. This Interactive is establish in the Physics Interactives section of our website and allows a learner to apply the skill of calculating areas and relating them to deportation values for a 2-stage rocket.