A particle is moving such that its position vector, r metres, at time t seconds, is given by r = e^t ext{cos} ext{i} + e^t ext{sin} ext{j} Show that the magnitude of the acceleration of the particle, a ext{ } ms}^{-2} ext{, is given by} a = 2e^t Fully justify your answer. - AQA - A-Level Maths Pure - Question 17 - 2022 - Paper 2

Next, differentiate the velocity to find the acceleration vector:

$a = \frac{d^2r}{dt^2} = \frac{d}{dt} \left( (e^t \text{cos} - e^t \text{sin}) ext{i} + (e^t \text{sin} + e^t \text{cos}) ext{j} \right)$

Applying the product rule again,

$a = (e^t \text{cos} + e^t \text{sin}) ext{i} + (e^t \text{sin} - e^t \text{cos}) ext{j}$

To find the magnitude of the acceleration, we compute:

$|a| = \sqrt{(2e^t)^2 + (e^t \text{sin})^2}$

This simplifies to:

$|a| = \sqrt{4e^{2t}} = 2e^t$

Hence, we have proved that the magnitude of the acceleration vector is indeed given by:

$a = 2e^t$