Edexcel IGCSE Physics: Double Science 复习笔记: 1.1.6 Area under a Velocity-Time Graph

Edexcel IGCSE Physics: Double Science 复习笔记: 1.1.6 Area under a Velocity-Time Graph

Area under a Velocity-Time Graph

 

  • The area under a velocity-time graph represents the displacement (or distance travelled) by an object

5.6.12-Velocity-Time-Area-graph

The displacement, or distance travelled, is represented by the area beneath the graph

  • If the area beneath the graph forms a triangle (i.e. the object is accelerating or decelerating), then the area can be determined by using the following formula:

 

Area = ½ × Base × Height

 

  • If the area beneath the graph forms a rectangle (i.e. the object is moving at a constant velocity), then the area can be determined by using the following formula:

 

Area = Base × Height

 

 

Determining Distance from a Velocity-Time Graph

  • Enclosed areas under velocity-time graphs represent total displacement (or total distance travelled)

 

5.6.12-Determining-Distance-on-a-V-T-graph 

Three enclosed areas (two triangles and one rectangle) under this velocity-time graph represents the total distance travelled

 

  • If an object moves with constant acceleration, its velocity-time graph will comprise of straight lines
    • In this case, calculate the distance travelled by working out the area of enclosed rectangles and triangles as in the image above

     

 

 

Worked Example

The velocity-time graph below shows a car journey which lasts for 160 seconds.2.1.8-Area-Under-a-V-T-graph-questionCalculate the total distance travelled by the car on this journey.

 

Step 1: Recall that the area under a velocity-time graph represents the distance travelled

    • In order to calculate the total distance travelled, the total area underneath the line must be determined

     

 

Step 2: Identify each enclosed area

 

    • In this example, there are five enclosed areas under the line
    • These can be labelled as areas 1, 2, 3, 4 and 5, as shown in the image below:

     

2.1.8-Area-Under-a-V-T-graph-solution

Step 3: Calculate the area of each enclosed shape under the line

    • Area 1 = area of a triangle = ½ × base × height = ½ × 40 × 17.5 = 350 m
    • Area 2 = area of a rectangle = base × height = 30 × 17.5 = 525 m
    • Area 3 = area of a triangle = ½ × base × height = ½ × 20 × 7.5 = 75 m
    • Area 4 = area of a rectangle = base × height = 20 × 17.5 = 350 m
    • Area 5 = area of a triangle = ½ × base × height = ½ × 70 × 25 = 875 m

     

 

Step 4: Calculate the total distance travelled by finding the total area under the line

 

    • Add up each of the five areas enclosed:

     

 

total distance = 350 + 525 + 75 + 350 + 875

 

total distance = 2175 m

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