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Figure 1 shows the speed-time graph of a cyclist moving on a straight road over a 7 s period - Edexcel - A-Level Maths Mechanics - Question 1 - 2006 - Paper 1

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Figure 1 shows the speed-time graph of a cyclist moving on a straight road over a 7 s period. The sections of the graph from t = 0 to t = 3, and from t = 3 to t = 7,... show full transcript

Worked Solution & Example Answer:Figure 1 shows the speed-time graph of a cyclist moving on a straight road over a 7 s period - Edexcel - A-Level Maths Mechanics - Question 1 - 2006 - Paper 1

Step 1

a) the graph from t = 0 to t = 3 is a straight line

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Answer

The graph from t = 0 to t = 3 being a straight line indicates that the cyclist is experiencing constant acceleration during this interval. Since speed is increasing uniformly, the cyclist is not speeding up or slowing down erratically.

Step 2

b) the graph from t = 3 to t = 7 is parallel to the t-axis

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Answer

The section of the graph from t = 3 to t = 7 being parallel to the t-axis indicates that the cyclist is traveling at a constant speed during this time period, meaning there is no acceleration.

Step 3

c) Find the distance travelled by the cyclist during this 7 s period

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Answer

To calculate the distance travelled by the cyclist, we can break it down into two parts:

  1. From t = 0 to t = 3 seconds: The speed increases from 0 to 5 m/s. The average speed for this period can be calculated as:

    Average Speed=0+52=2.5 m/s\text{Average Speed} = \frac{0 + 5}{2} = 2.5 \text{ m/s}

    The distance covered in this interval is:

    Distance=Average Speed×Time=2.5 m/s×3 s=7.5 m\text{Distance} = \text{Average Speed} \times \text{Time} = 2.5 \text{ m/s} \times 3 \text{ s} = 7.5 \text{ m}

  2. From t = 3 to t = 7 seconds: The speed is constant at 5 m/s for 4 seconds. The distance covered in this interval is:

    Distance=Speed×Time=5 m/s×4 s=20 m\text{Distance} = \text{Speed} \times \text{Time} = 5 \text{ m/s} \times 4 \text{ s} = 20 \text{ m}

The total distance travelled during the entire 7 seconds is:

Total Distance=7.5 m+20 m=27.5 m\text{Total Distance} = 7.5 \text{ m} + 20 \text{ m} = 27.5 \text{ m}

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