1. A particle P moves with constant acceleration (2i - 3j) ms⁻² At time t = 0, P is moving with velocity 4i ms⁻¹ (a) Find the velocity of P at time t = 2 seconds - Edexcel - A-Level Maths Mechanics - Question 1 - 2021 - Paper 1

To find the position vector of particle P at time t = 3 seconds, we use the following formula:

$r = r_0 + ut + \frac{1}{2}at^2$

Where:

• $r$ is the position vector,
• $r_0$ is the initial position vector,
• $u$ is the initial velocity,
• $a$ is the acceleration,
• $t$ is the time.

Given:

• Initial position vector, $r_0 = (i + j)$ m
• Initial velocity, $u = 4i$ ms⁻¹
• Acceleration, $a = (2i - 3j)$ ms⁻²
• Time, $t = 3$ seconds

Now, substituting these values into the formula:

1. Calculate the initial term:

$r_0 + ut = (i + j) + (4i)(3) = (i + j) + 12i = 13i + j$

2. Calculate the second term:

$\frac{1}{2}at^2 = \frac{1}{2}(2i - 3j)(3^2) = \frac{1}{2}(2i - 3j)(9) = (9i - 13.5j)$

Now, combine both parts:

$r = (13i + j) + (9i - 13.5j)$ $r = (13i + 9i) + (j - 13.5j)$ $r = 22i - 12.5j$

Thus, the position vector of P relative to O at time t = 3 seconds is: $r = 22i - 12.5j ext{ m}$