Velocity-Time Graphs (WJEC GCSE Science (Double Award)): Revision Note

A cyclist is training for a cycling tournament.

The velocity-time graph below shows the cyclist's motion as they cycle along a flat, straight road.

(a) In which section (A, B, C, D, or E) of the velocity-time graph is the cyclist's acceleration the largest?

(b) Calculate the cyclist's acceleration between 5 and 10 seconds. 

Answer:

Part (a)

Step 1: Recall that the slope of a velocity-time graph represents the magnitude of acceleration

• The slope of a velocity-time graph indicates the magnitude of acceleration

  Therefore, the only sections of the graph where the cyclist is accelerating are sections B and D

• Sections A, C, and E are flat; in other words, the cyclist is moving at a constant velocity (therefore, not accelerating)

Step 2: Identify the section with the steepest slope

• Section D of the graph has the steepest slope

• Hence, the largest acceleration is shown in section D 

Part (b)

Step 1: Recall that the gradient of a velocity-time graph gives the acceleration

• Calculating the gradient of a slope on a velocity-time graph gives the acceleration for that time period

Step 2: Draw a large gradient triangle at the appropriate section of the graph

• A gradient triangle is drawn for the time period between 5 and 10 seconds 

Step 3: Calculate the size of the gradient and state this as the acceleration

• The acceleration is given by the gradient, which can be calculated using:

• Therefore, the cyclist accelerated at 1 m/s² between 5 and 10 seconds

Use the entire slope, where possible, to calculate the gradient. Examiners tend to award credit if they see a large gradient triangle used.

Remember to actually draw the lines directly on the graph itself, particularly when the question asks you to use the graph to calculate the acceleration. 

The velocity-time graph below shows a car journey which lasts for 160 seconds.

Calculate the total distance travelled by the car on this journey.

Answer:

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

• 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:

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