At time t seconds a particle, P, has position vector r metres, with respect to a fixed origin, such that $$ r = (3t^2 - 5t) extbf{i} + (8t - t^2) extbf{j} $$ 14 (a) Find the exact speed of P when t = 2. 14 (b) Bella claims that the magnitude of acceleration of P will never be zero. Determine whether Bella's claim is correct. Fully justify your answer.

A-Level AQA Maths Mechanics Variable Acceleration - 2D: At time t seconds a particle, P,

At time t seconds a particle, P, has position vector r metres, with respect to a fixed origin, such that $$ r = (3t^2 - 5t) extbf{i} + (8t - t^2) extbf{j} $$ 14 (a) Find the exact speed of P when t = 2 - AQA - A-Level Maths Mechanics - Question 14 - 2020 - Paper 2

Question 14

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At time t seconds a particle, P, has position vector r metres, with respect to a fixed origin, such that $$ r = (3t^2 - 5t) extbf{i} + (8t - t^2) extbf{j} $$ 14 (... show full transcript

Worked Solution & Example Answer:At time t seconds a particle, P, has position vector r metres, with respect to a fixed origin, such that $$ r = (3t^2 - 5t) extbf{i} + (8t - t^2) extbf{j} $$ 14 (a) Find the exact speed of P when t = 2 - AQA - A-Level Maths Mechanics - Question 14 - 2020 - Paper 2

Step 1

Find the exact speed of P when t = 2

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Answer

To find the speed of the particle P at time t = 2, we first need to determine the velocity vector v by differentiating the position vector r with respect to time t.

Differentiate the position vector:

$v = \frac{dr}{dt} = \frac{d}{dt}((3t^2 - 5t)\textbf{i} + (8t - t^2)\textbf{j})$

This yields:

$v = (6t - 5)\textbf{i} + (8 - 2t)\textbf{j}$
Substitute t = 2 into the velocity vector:

$v = (6(2) - 5)\textbf{i} + (8 - 2(2))\textbf{j}$

Simplifying gives:

$v = (12 - 5)\textbf{i} + (8 - 4)\textbf{j} = 7\textbf{i} + 4\textbf{j}$
Calculate the magnitude of the velocity vector to determine the speed:

$\text{Speed} = |v| = \sqrt{(7)^2 + (4)^2} = \sqrt{49 + 16} = \sqrt{65} = 4\sqrt{5} \text{ m/s}$

Thus, the exact speed of P when t = 2 is $4\sqrt{5} \text{ m/s}$ .

Step 2

Determine whether Bella's claim is correct.

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Answer

To analyze Bella's claim that the magnitude of acceleration of P will never be zero, we first need to calculate the acceleration vector.

Differentiate the velocity vector to find the acceleration:

$a = \frac{dv}{dt} = \frac{d}{dt}((6t - 5)\textbf{i} + (8 - 2t)\textbf{j})$

This results in:

$a = (6)\textbf{i} + (-2)\textbf{j}$
To check if the magnitude of the acceleration can be zero, we calculate:

$|a| = \sqrt{(6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} \neq 0$
Since the values are constant irrespective of t, the magnitude of the acceleration will always be:

$|a| = \sqrt{40} = 2\sqrt{10} \text{ which is clearly } \geq 2$

Therefore, Bella's claim that the magnitude of acceleration will never be zero is correct.

At time t seconds a particle, P, has position vector r metres, with respect to a fixed origin, such that $$ r = (3t^2 - 5t) extbf{i} + (8t - t^2) extbf{j} $$ 14 (a) Find the exact speed of P when t = 2 - AQA - A-Level Maths Mechanics - Question 14 - 2020 - Paper 2

Worked Solution & Example Answer:At time t seconds a particle, P, has position vector r metres, with respect to a fixed origin, such that $$ r = (3t^2 - 5t) extbf{i} + (8t - t^2) extbf{j} $$ 14 (a) Find the exact speed of P when t = 2 - AQA - A-Level Maths Mechanics - Question 14 - 2020 - Paper 2

Find the exact speed of P when t = 2

Determine whether Bella's claim is correct.

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