Kinematical equation by graph
Answers
Consider an object that moves with a constant velocity of +5 m/s for a time period of 5 seconds and then accelerates to a final velocity of +15 m/s over the next 5 seconds. Such a verbal description of motion can be represented by a velocity-time graph. The graph is shown below.
The horizontal section of the graph depicts a constant velocity motion, consistent with the verbal description. The positively sloped (i.e., upward sloped) section of the graph depicts a positive acceleration, consistent with the verbal description of an object moving in the positive direction and speeding up from 5 m/s to 15 m/s. The slope of the line can be computed using the rise over run ratio. Between 5 and 10 seconds, the line rises from 5 m/s to 15 m/s and runs from 5 s to 10 s. This is a total rise of +10 m/s and a total run of 5 s. Thus, the slope (rise/run ratio) is (10 m/s)/(5 s) = 2 m/s2. Using the velocity-time graph, the acceleration of the object is determined to be 2 m/s2 during the last five seconds of the object's motion. The displacement of the object can also be determined using the velocity-time graph. The area between the line on the graph and the time-axis is representative of the displacement; this area assumes the shape of a trapezoid. As discussed in Lesson 4, the area of a trapezoid can be equated to the area of a triangle lying on top of the area of a rectangle
