A particle is moving in a straight line with velocity $v$ m s$^{-1}$ at time $t$ seconds as shown by the graph below. Use the trapezium rule with four strips to estimate the distance travelled by the particle during the time period $20 \leq t \leq 100$. Over the same time period, the curve can be very closely modelled by a particular quadratic. Explain how you could find an alternative estimate using this quadratic.

A-Level AQA Maths Pure Quadratics: A particle is moving in a straig

A particle is moving in a straight line with velocity $v$ m s$^{-1}$ at time $t$ seconds as shown by the graph below - AQA - A-Level Maths Pure - Question 15 - 2020 - Paper 2

Question 15

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A particle is moving in a straight line with velocity $v$ m s$^{-1}$ at time $t$ seconds as shown by the graph below. Use the trapezium rule with four strips to est... show full transcript

Worked Solution & Example Answer:A particle is moving in a straight line with velocity $v$ m s$^{-1}$ at time $t$ seconds as shown by the graph below - AQA - A-Level Maths Pure - Question 15 - 2020 - Paper 2

Step 1

Use the trapezium rule with four strips to estimate the distance travelled by the particle during the time period 20 ≤ t ≤ 100

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Answer

To apply the trapezium rule, we first need to determine the values of the velocity at the given time intervals.

Determine the strip width: The time interval from 20 to 100 is $h = \frac{100 - 20}{4} = 20$ seconds.
Identify the corresponding $y$ values:
- When $t = 20$ , $v = 131$ m/s
- When $t = 40$ , $v = 140$ m/s
- When $t = 60$ , $v = 120$ m/s
- When $t = 80$ , $v = 80$ m/s
- When $t = 100$ , $v = 0$ m/s
Apply the trapezium rule:

$ext{Area} = \frac{h}{2} (y_0 + 2y_1 + 2y_2 + 2y_3 + y_4)$

Where:
- $y_0 = 131$ , $y_1 = 140$ , $y_2 = 120$ , $y_3 = 80$ , $y_4 = 0$ .
- Substituting these values gives:
$ext{Area} = \frac{20}{2} (131 + 2 \times 140 + 2 \times 120 + 2 \times 80 + 0)$

Calculating this results in:

$\text{Area} = 10 (131 + 280 + 240 + 160) = 10 \times 811 = 8110 \text{ meters}$

Therefore, the estimated distance travelled by the particle is 8110 meters.

Step 2

Explain how you could find an alternative estimate using this quadratic

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Answer

To find an alternative estimate using the quadratic model, you would follow these steps:

Formulate the quadratic equation that closely models the curve derived from the velocity graph over the range of interest. The general form of a quadratic equation is:

$y = ax^2 + bx + c$

where $a$ , $b$ , and $c$ are constants determined from the known data points.
Integrate the quadratic function between the limits of integration 20 and 100 seconds:

$\text{Distance} = \int_{20}^{100} (ax^2 + bx + c) \, dx$
Evaluate the integral to find the total distance travelled over the time period. This method often provides a more accurate estimate compared to the trapezium rule, especially when the data can be well-represented by a continuous function.

A particle is moving in a straight line with velocity $v$ m s$^{-1}$ at time $t$ seconds as shown by the graph below - AQA - A-Level Maths Pure - Question 15 - 2020 - Paper 2

Worked Solution & Example Answer:A particle is moving in a straight line with velocity $v$ m s$^{-1}$ at time $t$ seconds as shown by the graph below - AQA - A-Level Maths Pure - Question 15 - 2020 - Paper 2

Use the trapezium rule with four strips to estimate the distance travelled by the particle during the time period 20 ≤ t ≤ 100

Explain how you could find an alternative estimate using this quadratic

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