A-Level AQA Maths Mechanics Working with Vectors: 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 $$ r = (3t^2 - 5t) oldsymbol{i} + (8t - t^2) oldsymbol{j} $$ 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 - AQA - A-Level Maths Mechanics - Question 14 - 2020 - Paper 2

Question 14

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

At time t seconds a particle, P, has position vector r metres, with respect to a fixed origin, such that $$ r = (3t^2 - 5t) oldsymbol{i} + (8t - t^2) oldsymbol{j... 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 $$ r = (3t^2 - 5t) oldsymbol{i} + (8t - t^2) oldsymbol{j} $$ 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 - 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 determine the speed of particle P at time t = 2, we first need to find the velocity vector, v, by differentiating the position vector, r.

Differentiating r with respect to time t:

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

Calculating the derivatives:

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

Now substituting t = 2 into the velocity vector:

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

$= (12 - 5) \boldsymbol{i} + (8 - 4) \boldsymbol{j}$

$= 7 \boldsymbol{i} + 4 \boldsymbol{j}$

To find the speed, we calculate the magnitude of the velocity vector:

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

Thus, the exact speed of P when t = 2 is:

$\sqrt{65} \, \text{m s}^{-1}$

Step 2

Determine whether Bella's claim is correct.

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Answer

To check Bella's claim about the acceleration of P, we start by finding the acceleration vector a, which is the derivative of the velocity vector v:

$a = \frac{dv}{dt}$

Differentiating the velocity vector:

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

Calculating the derivatives, we obtain:

$a = (6) \boldsymbol{i} + (-2) \boldsymbol{j}$

The magnitude of the acceleration vector is given by:

$|a| = \sqrt{(6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40}$

Since the magnitude of acceleration does not depend on time t, we have:

$|a| = \sqrt{40} \, \text{which is always positive.}$

Thus, for all values of t, the magnitude of the acceleration will never be zero. Therefore, Bella's claim is correct.

Find the exact speed of P when t = 2

Determine whether Bella's claim is correct.

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