A15: Gradients & Areas Under Graphs

Find the gradient between points (1, 3) and (4, 9) on a graph.

Solution:

Gradient = ^{9 - 3}⁄_{4 - 1} = ⁶⁄₃ = 2

A distance-time graph shows a car travelling from 0 km to 150 km in 3 hours. Calculate the speed.

Solution:

Speed = gradient = ^{150 - 0}⁄_{3 - 0} = 50 km/h

At a specific point on a distance-time graph, the gradient is 25. What does this mean?

Solution:

The instantaneous speed is 25 (units per hour/minute depending on scale).

A velocity-time graph shows velocity increasing from 0 m/s to 20 m/s in 5 seconds. Calculate acceleration.

Solution:

Acceleration = gradient = ^{20 - 0}⁄_{5 - 0} = 4 m/s²

On a velocity-time graph, what does a horizontal line mean?

Solution:

Gradient = 0, so acceleration = 0. The object is moving at constant velocity.

Find the area under the line y = 2x from x = 0 to x = 4.

Solution:

This forms a triangle with base 4 and height 8.

Area = ½ × 4 × 8 = 16 square units

A car accelerates uniformly from 0 to 15 m/s in 10 seconds, then travels at constant speed for 20 seconds. Find total distance.

Solution:

Part 1 (triangle): Area = ½ × 10 × 15 = 75 m

Part 2 (rectangle): Area = 20 × 15 = 300 m

Total distance: 75 + 300 = 375 m

A velocity-time graph shows:

• 0-5s: velocity increases from 0 to 10 m/s
• 5-15s: constant velocity of 10 m/s
• 15-20s: velocity decreases from 10 m/s to 0

Calculate total distance.

Solution:

Area 1 = ½ × 5 × 10 = 25 m

Area 2 = 10 × 10 = 100 m

Area 3 = ½ × 5 × 10 = 25 m

Total: 25 + 100 + 25 = 150 m

Estimate the area under a curve by counting squares. If 15 full squares and 8 half-squares are under the curve, and each square represents 2 units²:

Solution:

Total squares = 15 + 4 = 19 squares

Area ≈ 19 × 2 = 38 units²