At time $t$ seconds, where $t > 0$, a particle $P$ moves so that its acceleration $a ext{ m s}^{-2}$ is given by a = (1-4t)i + (3-t)j At the instant when $t = 0$, the velocity of $P$ is 36 ms$^{-1}$ (a) Find the velocity of $P$ when $t = 4$ (b) Find the value of $t$ at the instant when $P$ is moving in a direction perpendicular to $i$ (ii) At time $t$ seconds, where $t > 0$, a particle $Q$ moves so that its position vector $r$ metres, relative to a fixed origin $O$, is given by r = (t^2 - t)i + 3j Find the value of $t$ at the instant when the speed of $Q$ is 5 ms$^{-1}$. - Edexcel - A-Level Maths Mechanics - Question 3 - 2020 - Paper 1

To find the velocity of particle $P$, we first integrate the acceleration vector to get the velocity vector.

The acceleration vector is: $a = (1 - 4t)i + (3 - t)j$

Integrating each component with respect to $t$ gives us: $v = \int a \, dt = \int ((1 - 4t)i + (3 - t)j) \, dt = (t - 2t^2)i + \left(3t - \frac{1}{2}t^2\right)j + C$

Given that at $t = 0$, the velocity $v = 36j$, we find:

i-component: $0 = 0 + C_1$ implies $C_1 = 0$

j-component: $36 = 0 + C_2$ implies $C_2 = 36$

Thus, the velocity vector becomes: $v = (t - 2t^2)i + \left(3t - \frac{1}{2}t^2 + 36\right)j$

Now substituting $t = 4$: $v(4) = (4 - 2 * 4^2)i + \left(3 * 4 - \frac{1}{2} * 16 + 36\right)j$

Calculating further results in: $v(4) = (4 - 32)i + (12 - 8 + 36)j = -28i + 40j$