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A car's motion is represented by a distance-time graph. Describe the motion in segments OA, AB, and BC, and calculate the average speed for the first 10 seconds.

Physics

ScanSolve AI Answer

Answer

a(ii)

5 m/s\text{5 m/s}

Step-by-step solution

3 steps

Part a(i)

  • OA: The gradient of the distance-time graph is increasing, indicating that the car is accelerating (increasing its speed).
  • AB: The gradient is constant and positive, meaning the car is moving at a constant speed.
  • BC: The gradient is decreasing and eventually becomes zero, which means the car is decelerating (decreasing its speed) and then stops.

Part a(ii)

5 m/s\text{5 m/s}

  1. Identify the total distance covered in the first 10 seconds.

    From the graph, at t=0 st = 0\,\text{s}, distance =0 m= 0\,\text{m}.
    At t=10 st = 10\,\text{s}, distance =50 m= 50\,\text{m}.
    Total distance =50 m−0 m=50 m= 50\,\text{m} - 0\,\text{m} = 50\,\text{m}.

  2. Identify the total time taken.

    Total time =10 s−0 s=10 s= 10\,\text{s} - 0\,\text{s} = 10\,\text{s}.

  3. Calculate the average speed.

    Average speed =Total distanceTotal time= \frac{\text{Total distance}}{\text{Total time}}

    Average speed=50 m10 s\text{Average speed} = \frac{50\,\text{m}}{10\,\text{s}} Average speed=5 m/s\text{Average speed} = 5\,\text{m/s}

    The car's average speed during the first 10s is 5 m/s\boxed{\text{5 m/s}}.

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