[In this question, the unit vectors i and j are due east and due north respectively:] A particle P is moving with constant velocity (−5i + 8j) m s⁻¹ - Edexcel - A-Level Maths Mechanics - Question 6 - 2008 - Paper 1

To determine the direction, we can find the angle that the velocity vector makes with the north direction (the positive j-axis).

Using the formula for the angle,

$\theta = \tan^{-1}\left(\frac{8}{-5}\right).$

Calculating this gives us an angle in the fourth quadrant. To convert this to a bearing:

1. Compute the angle:

   $\theta \approx \tan^{-1}\left(-1.6\right) \approx -58.0^{\circ}$.

2. The bearing is given by:

   $\text{Bearing} = 360 - 58.0 = 328^{\circ}.$

At time t = 3 s, the position vector of P is given by:

$\text{Position} = (7i - 10j) + 3(-5i + 8j) = (7 - 15)i + (-10 + 24)j = -8i + 14j.$

At this moment, the velocity vector is (u + iv), leading to:

$\text{Change in Position} = (-8i + 14j) = 3(-5i + 8j) + 3(u + iv).$

By equating components, we can find u and v:

1. For the i-component:

   $-8 = 3u - 15 \rightarrow u = 2.$

2. For the j-component:

   $14 = 24 + 3v \rightarrow v = -3.$

The position due south of A can be represented as (7i - k), where k is the y-component, which can be solved with:

The initial point A is at (7i - 10j) and after 3 seconds it moves with a new velocity:

$\text{New Position} = (7 - 8) + (-10 + 14) = -1i + 14j.$

The particle continues to move in the direction of velocity (2i - 3j) m/s.

To calculate the time taken to reach the south position, let's set this equality and solve for time t:

$\text{Position after } t ext{ seconds} = 7i - (10 + 3t) j.$

Set the x-coordinate to 0 to find time:

$7 - 2t = 0 \rightarrow t = 3.5 \, s.$

Total time taken = time at the start + time to reach = 3 s + 3.5 s = 10.5 s.