A particle P moves with constant acceleration (2i - 3j) m s^-2 At time t = 0, P is moving with velocity 4i m s^-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 P relative to O at time $t = 3$ seconds, we use the equation:

$r = (i + j) + ut + \frac{1}{2}at^2$

where:

• Initial position vector, $r_0 = (i + j)$ m
• Initial velocity, $u = 4i$ m/s
• Acceleration, $a = (2i - 3j)$ m/s²
• Time, $t = 3$ seconds.

First, we calculate:

• The term $ut$:
  $ut = (4i) \times 3 = 12i$

• The term rac{1}{2}at^2:
  $\frac{1}{2}(2i - 3j) \times (3)^2 = (i - \frac{3}{2}j) \times 9 = 9i - \frac{27}{2}j$

Now substituting back into the position formula:

$r = (i + j) + 12i + (9i - \frac{27}{2}j)$

Combining the vectors: $r = (1 + 12 + 9)i + (1 - \frac{27}{2})j$

Simplifying further: $r = 22i - 12.5j$

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