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

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

1 Answer

6 votes

Answer:

See below ~

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
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