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 $20 \, s \leq t \leq 100 \, s$. 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 Mechanics Kinematics Graphs: 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 Mechanics - 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 Mechanics - 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 identify the interval and the corresponding $y$ -values based on the graph:

We divide the interval [ $20, 100$ ] into four equal strips, giving us a strip width of: $h = \frac{100 - 20}{4} = 20$
The $x$ -values (time) are: $20, 40, 60, 80, 100$ .
The corresponding $y$ -values (velocity) at these points are:
- $y_0 = 120$ (at $t=20$ )
- $y_1 = 140$ (at $t=40$ )
- $y_2 = 80$ (at $t=60$ )
- $y_3 = 20$ (at $t=80$ )
- $y_4 = 0$ (at $t=100$ )

Using the trapezium rule formula:

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

Substituting the values:

$\text{Area} \approx \frac{20}{2} (120 + 2(140) + 2(80) + 2(20) + 0)$ $= 10 (120 + 280 + 160 + 40)$ $= 10 \times 600 = 6000$

Thus, the estimated distance travelled by the particle from $20$ s to $100$ s is $6000$ 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, we could follow these steps:

Identify the quadratic function that closely models the velocity data over the interval $20 \leq t \leq 100$ .
Integrate this quadratic function over the interval from $20$ to $100$ to find the total distance travelled.
The integral of the quadratic function will provide a more accurate estimate of the distance compared to the trapezium rule, as it accounts for the curvature of the velocity graph.
Use numerical methods or symbolic integration to evaluate the integral, specifically:
$\int_{20}^{100} f(t) \, dt$ , where $f(t)$ is the quadratic 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 Mechanics - 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 Mechanics - 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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