Lesson 4: Describing Motion with Velocity vs. Time Graphs

Determining the Area on a v-t Graph

As you learned in an earlier section of this lesson, a plot of velocity vs. time can be used to determine the acceleration of an object (slope = acceleration). In this part of the lesson, you will learn how a plot of velocity vs. time can also be used to determine the distance traveled by an object. For velocity vs. time graphs, the area bounded by the line and the axes represents the distance traveled.

The diagram below shows three different velocity-time graphs; the shaded regions between the line and the axes represent the distance traveled during the stated time interval.

The shaded area is representative of the distance traveled by the object during the time interval from 0 seconds to 6 seconds. This representation of the distance traveled takes on the shape of a rectangle whose area can be calculated using the appropriate equation.
The shaded area is representative of the distance traveled by the object during the time interval from 0 seconds to 4 seconds. This representation of the distance traveled takes on the shape of a triangle whose area can be calculated using the appropriate equation.
The shaded area is representative of the distance traveled by the object during the time interval from 2 seconds to 5 seconds. This representation of the distance traveled takes on the shape of a trapezoid whose area can be calculated using the appropriate equation.

The method used to find the area under a line on a velocity-time graph depends on whether the section bounded by the line and the axes is a rectangle, a triangle or a trapezoid. Area formulae for each shape are given below.

Calculating the Area of a Rectangle

The shaded rectangle on the velocity-time graph, below, has a base of 6 s and a height of 30 m/s.

Area of rectangle: A = b x h = (6 s) x (30 m/s) = 180 m.

The object was displaced 180 meters during the first 6 seconds of motion.

Area = b * h Area = (6 s) * (30 m/s) Area = 180 m

Now check your understanding by finding the distance traveled by the object in each of the following cases.

Depress mouse to see answer to Practice A. The area is b x h where b = 4 s and h = 30 m/s. Thus, the area is 120 m (A = 4 s x 30 m/s). That is, the object was displaced 120 m during the first 4 seconds of motion.

Depress mouse to see answer to Practice B. The area is b x h where b = 3 s and h = 30 m/s. Thus, the area is 90 m (A = 3s x 30 m/s). That is, the object was displaced 90 m during the time interval from 3 to 6 seconds.

Calculating the Area of a Triangle

The shaded triangle on the velocity-time graph, below, has a base of 4 seconds and a height of 40 m/s.

Area of triangle: A = 0.5 * b * h = (0.5) * (4 s) * (40 m/s) = 80 m.

The object was displaced 80 meters during the first four seconds of motion.

Area = 0.5 * b * h Area = (0.5) * (4 s) * (40 m/s) Area = 80 m

Check your understanding by finding the distance traveled by the object in each of the following cases.

Depress mouse to see answer to Practice A. The area is 0.5 x b x h where b = 1 s and h = 10 m/s. Thus, the area is 5 m (A = 0.5 x 1 s x 10 m/s). That is, the object was displaced 5 m during the first second of motion.

Depress mouse to see answer to Practice B. The area is 0.5 x b x h where b = 3 s and h = 30 m/s. Thus, the area is 45 m (A = 0.5 x 3 s x 30 m/s). That is, the object was displaced 45 m during the first 3 seconds of motion.

Calculating the Area of a Trapezoid

The shaded trapezoid on the velocity-time graph, below, has a base of 2 seconds and heights of 10 m/s (on the left side) and 30 m/s (on the right side).

Area of trapezoid:
   A = (0.5) * (b) * (h1 + h2)
      = (0.5) * (2 s) * (10 m/s + 30 m/s) = 40 m.

The object was displaced 40 meters during the time interval from 1 second to 3 seconds.

Area = 0.5 * b * (h1 + h2) Area = (0.5) * (2 s) * (10 m/s + 30 m/s) Area = 40 m

Now check your understanding by finding the distance traveled by the object in each of the following cases.

Depress mouse to see answer to Practice A. The area is 0.5 * b * (h1 + h2) where b = 1 s and h1 = 20 m/s and h2 = 30 m/s. Thus, the area is 25 m [A = 0.5 * 1 s * (20 m/s + 30 m/s)]. That is, the object was displaced 25 m during the time interval from 2 to 3 seconds.

Depress mouse to see answer to Practice B. The area is 0.5 * b * (h1 + h2) where b = 2 s and h1 = 30 m/s and h2 = 10 m/s. Thus, the area is 40 m [A = 0.5 * 2 s * (30 m/s + 10 m/s)]. That is, the object was displaced 40 m during the first 2 seconds of motion.

Alternative Method for Calculating the Area of a Trapezoid

An alternative method of determining the area of a trapezoid involves breaking the trapezoid into a triangle and a rectangle. The areas of the triangle and rectangle are computed individually; the area of the trapezoid is then the sum of the areas of the triangle and the rectangle. This method is illustrated below.

Triangle: Area = (0.5) * (2 s) * ( 20 m/s) = 20 m

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

Trapezoid: Area = 20 m + 20 m = 40 m

To review, in this lesson you learned that the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an the object during that time interval. The shaded region can be identified as either a rectangle, triangle, or trapezoid whose area can subsequently be determined using the appropriate formula. Once calculated, this area represents the displacement of the object during the time period indicated.

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