Velocity-Time Graphs (WJEC GCSE Physics)

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

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