A particle moves with acceleration a = (4−t) ext m/s2. Its initial velocity is v0 = 3 ext m/s. Find its velocity, displacement, and time to attain maximum velocity.

Step 1: Understand the given information.
The acceleration of the particle is given by $a = (4-t) \text{ m/s}^2$.
The initial velocity (at $t=0$) is $v_0 = 3 \text{ m/s}$.

a) (i) Express the velocity $v$ in terms of $t$.
Velocity is the integral of acceleration with respect to time: $v = \int a\,dt$.

$$v = \int (4-t)\,dt$$ $$v = 4t - \frac{t^2}{2} + C$$

To find the constant of integration $C$, use the initial condition: when $t=0$, $v=3$.

$$3 = 4(0) - \frac{(0)^2}{2} + C$$ $$3 = C$$

Substitute $C=3$ back into the velocity equation:

$$v = 4t - \frac{t^2}{2} + 3$$

The velocity $v$ in terms of $t$ is $\boxed{v = 4t - \frac{t^2}{2} + 3 \text{ m/s}}$.

a) (ii) Find the velocity of the particle after 3 seconds.
Substitute $t=3$ into the velocity equation:

$$v = 4(3) - \frac{(3)^2}{2} + 3$$ $$v = 12 - \frac{9}{2} + 3$$ $$v = 15 - 4.5$$ $$v = 10.5 \text{ m/s}$$

The velocity after 3 seconds is $\boxed{10.5 \text{ m/s}}$.

b) (i) Calculate the time taken to attain maximum velocity.
Maximum velocity occurs when the acceleration is zero ($a=0$).

$$a = 4-t$$

Set $a=0$:

$$0 = 4-t$$ $$t = 4 \text{ s}$$

The time taken to attain maximum velocity is $\boxed{4 \text{ s}}$.

b) (ii) Calculate the displacement of the particle by the time it attains the maximum velocity.
Displacement $s$ is the integral of velocity with respect to time: $s = \int v\,dt$.
We use the velocity equation $v = 4t - \frac{t^2}{2} + 3$.

$$s = \int \left(4t - \frac{t^2}{2} + 3\right)\,dt$$ $$s = \frac{4t^2}{2} - \frac{t^3}{2 \cdot 3} + 3t + D$$ $$s = 2t^2 - \frac{t^3}{6} + 3t + D$$

Assume initial displacement is zero: when $t=0$, $s=0$.

$$0 = 2(0)^2 - \frac{(0)^3}{6} + 3(0) + D$$ $$0 = D$$

So, the displacement equation is:

$$s = 2t^2 - \frac{t^3}{6} + 3t$$

The particle attains maximum velocity at $t=4$ s. Substitute $t=4$ into the displacement equation:

$$s = 2(4)^2 - \frac{(4)^3}{6} + 3(4)$$ $$s = 2(16) - \frac{64}{6} + 12$$ $$s = 32 - \frac{32}{3} + 12$$ $$s = 44 - \frac{32}{3}$$

To combine these, find a common denominator:

$$s = \frac{44 \times 3}{3} - \frac{32}{3}$$ $$s = \frac{132 - 32}{3}$$ $$s = \frac{100}{3} \text{ m}$$

The displacement of the particle by the time it attains maximum velocity is $\boxed{\frac{100}{3} \text{ m}}$ or $\boxed{33\frac{1}{3} \text{ m}}$.