Displacement On Velocity Time Graph

Determining the Area on a v-t Graph

As learned in an earlier part of this lesson, a plot of velocity-fourth dimension can be used to determine the acceleration of an object (the gradient). In this part of the lesson, we will acquire how a plot of velocity versus fourth dimension can also exist used to determine the deportation of an object. For velocity versus time graphs, the area bound by the line and the axes represents the displacement. The diagram below shows three dissimilar velocity-fourth dimension graphs; the shaded regions between the line and the time-axis stand for the displacement during the stated fourth dimension interval.

The method used to notice the area nether 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. Surface area formulas for each shape are given below.

Rectangle	Triangle	Trapezoid
Area = b • h	Area = ½ • b • h	Surface area = ½ • b • (h_one + h₂)

Calculating the Area of a Rectangle

At present nosotros will wait at a few instance computations of the area for each of the higher up geometric shapes. First consider the adding of the area for a few rectangles. The solution for finding the expanse is shown for the first example below. The shaded rectangle on the velocity-time graph has a base of operations of 6 s and a height of xxx yard/s. Since the area of a rectangle is constitute by using the formula A = b ten h, the surface area is 180 chiliad (vi s x 30 grand/s). That is, the object was displaced 180 meters during the first half dozen seconds of motion.

Surface area = b * h
Expanse = (6 due south) * (30 yard/southward)

Area = 180 m

Now endeavor the following 2 practice problems as a cheque of your understanding. Make up one's mind the displacement (i.due east., the area) of the object during the first 4 seconds (Exercise A) and from iii to half-dozen seconds (Practice B).

Calculating the Expanse of a Triangle

Now we will look at a few example computations of the area for a few triangles. The solution for finding the area is shown for the first example below. The shaded triangle on the velocity-time graph has a base of 4 seconds and a height of forty m/s. Since the area of triangle is found past using the formula A = ½ * b * h, the expanse is ½ * (4 s) * (xl thousand/southward) = 80 yard. That is, the object was displaced eighty meters during the iv seconds of motion.

Area = ½ * b * h
Area = ½ * (4 s) * (forty chiliad/south)

Area = 80 m

Now effort the post-obit two practise bug as a check of your understanding. Determine the displacement of the object during the first second (Practise A) and during the showtime 3 seconds (Practice B).

Calculating the Area of a Trapezoid

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

Area = ½ * b * (h_i + h_two)
Area = ½ * (2 s) * (10 m/s + 30 one thousand/s)

Area = xl k

Now try the following two practise issues equally a bank check of your understanding. Determine the displacement of the object during the fourth dimension interval from 2 to 3 seconds (Practice A) and during the commencement 2 seconds (Exercise B).

Alternative Method for Trapezoids

An alternative means of determining the surface 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 then the sum of the areas of the triangle and the rectangle. This method is illustrated in the graphic below.

Triangle: Expanse = ½ * (2 s) * (20 m/s) = 20 thousand

Rectangle: Area = (2 s) * (10 m/s) = 20 m

Total Area = twenty m + twenty thou = 40 m

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 deportation of an object during that detail time period. The surface area can exist identified equally a rectangle, triangle, or trapezoid. The area can exist subsequently determined using the appropriate formula. Once calculated, this area represents the displacement of the object.

Investigate!

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

We Would Like to Suggest ...

Sometimes it isn't plenty to simply read nigh it. You have to interact with it! And that's exactly what y'all exercise when yous utilize one of The Physics Classroom'southward Interactives. We would like to suggest that you lot combine the reading of this page with the use of our Two Phase Rocket Interactive. This Interactive is found in the Physics Interactives section of our website and allows a learner to apply the skill of calculating areas and relating them to displacement values for a two-stage rocket.