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

To find the speed of P at t = 3, we use the velocity function:

$v(t) = (-2 + 2t)i + (7 - 3t)j$

Substituting t = 3 into the equation: $v(3) = (-2 + 6)i + (7 - 9)j = 4i - 2j$

To calculate the speed, we use the magnitude of velocity:

$|v| = \sqrt{(4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} ≈ 4.47 m/s$

To find the angle between the vector j (upwards) and the direction of motion of P at t = 3, we first identify the direction vector from our velocity:

$v(3) = 4i - 2j$

The direction can be represented as a right triangle, where:

• The opposite side (j component) = -2
• The adjacent side (i component) = 4

Calculating the angle θ using the tangent function: $\tan(θ) = \frac{2}{4} = \frac{1}{2}$

Thus, we find: $θ = \arctan\left( \frac{1}{2} \right)\approx 26.57°$

To find the angle relative to the vector j, we subtract from 90°: $90° - 26.57° = 63.43°$

Therefore, the angle between vector j and the direction of motion of P is 116.57° (acceptable to 117°).