Question

Q2 - Distance and average speed Use the graph to work out the following correct to 1 dp:

Answer 1

Answer:

See below ~

Step-by-step explanation:

Part 1 - Acceleration when t = 3

• Acceleration is constant during the time period t = 0 to t = 4
• a = change in velocity / change in time
• a = 22 - 11 / 4 - 0
• a = 11/4
• a = 2.75 m/s²

Part 2 - Deceleration when t = 12

• Deceleration is constant during the time period t = 8 to t = 15
• Distance covered in the time period = 1/2 x 7 x 22 = 77 m
• v² - u² = 2aS
• 0 - 484 = 2(77) × a
• a = -484/154
• a ≈ -3.14 m/s²

Part 3 - Total Distance Covered

• Area under the graph gives distance

1. From t = 0 to t = 4

• S = 1/2 x 11 x 4 + 11 x 4
• S = 22 + 44 = 66 m

    2. From t = 4 to t = 8

• S = 22 x 4 = 88 m

    3. From t = 8 to t = 15

• 77m (found in Part 2)

Total distance = 66 + 88 + 77 = 231 m

Part 4 - Average Speed

1. Avg. Speed (t = 0 to t =4)

• 66/4 = 16.5 m/s

    2. Avg. Speed (t = 4 to t = 8)

• 88/4 = 22 m/s

    3. Avg. Speed (t = 8 to t = 15)

• 77/7 = 11 m/s

Average Speed = 16.5 + 22 + 11 / 3

• 49.5/3
• 16.5 m/s