A particle P moves with constant acceleration (2i - 3j) m s² - Edexcel - A-Level Maths Mechanics - Question 5 - 2003 - Paper 1

Substituting t = 3 into the velocity equation:

$v = (-2i + j) + (2i - 3j)(3) = (-2 + 6)i + (1 - 9)j = 4i - 8j$

The speed |v| is given by:

$|v| = \sqrt{(4)^2 + (-8)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5} \approx 4.47 \text{ m s}^{-1}$

To find the angle between the vector j and the velocity vector v = 4i - 8j, we can use the arctangent function.

The direction of the velocity vector can be represented as:

$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{-8}{4} = -2$

Thus,

$\theta = \arctan(-2)$

To find the angle in relation to vector j (90 degrees), we get:

$\text{Angle} = 90° + \arctan(-2) \approx 116.6° \text{ (accept 117°)}$