A particle P starts from rest at a fixed point O and moves in a straight line. The velocity, $v\ \text{ms}^{-1}$, of the particle, $t\ \text{s}$ after passing through O is given by $v = 8t - ct^{3}$ where $c$ is a positive constant. The velocity of the particle is $12\ \text{ms}^{-1}$ after 2 seconds. Find

1. Substitute the given values of $v$ and $t$ into the velocity equation to find $c$

When $t = 2$, $v = 12$.
$12 = 8(2) - c(2)^3$
$12 = 16 - 8c$
$8c = 4$
$c = 0.5$

1. Find the expression for acceleration by differentiating velocity

From part (a), $c = 0.5$, so $v = 8t - 0.5t^3$.
$a = \frac{dv}{dt} = 8 - 1.5t^2$

2. Calculate the acceleration at $t = 2$

When $t = 2$,
$a = 8 - 1.5(2)^2$
$a = 8 - 1.5(4) = 8 - 6 = 2\ \text{ms}^{-2}$

1. Set velocity to zero to find when the particle changes direction

From part (a), $c = 0.5$, so $v = 8t - 0.5t^3$.
The particle changes direction when $v = 0$:
$8t - 0.5t^3 = 0$

2. Solve the equation for $t > 0$

$t(8 - 0.5t^2) = 0$
Since the particle starts at $t=0$, we look for $t > 0$:
$8 - 0.5t^2 = 0$
$0.5t^2 = 8$
$t^2 = 16$
$t = 4 \text{ (since } t > 0)$

1. Find the value of c from part (a)

Given $v = 12$ when $t = 2$:
$12 = 8(2) - c(2)^3$
$12 = 16 - 8c$
$8c = 4 \implies c = 0.5$
So, $v = 8t - 0.5t^3$.

2. Find the time when the particle changes direction from part (c)

Particle changes direction when $v = 0$:
$8t - 0.5t^3 = 0$
$t(8 - 0.5t^2) = 0$
Since $t > 0$, $0.5t^2 = 8 \implies t^2 = 16 \implies t = 4$.

3. Find the displacement function $s(t)$

$s = \int v \, dt = \int (8t - 0.5t^3) \, dt$
$s = 4t^2 - \frac{0.5t^4}{4} + C = 4t^2 - 0.125t^4 + C$
Since $s = 0$ when $t = 0$, $C = 0$.
$s = 4t^2 - 0.125t^4$

4. Calculate the distance travelled in the first 4 seconds

At $t = 4$:
$s(4) = 4(4)^2 - 0.125(4)^4 = 64 - 32 = 32\text{ m}$

5. Calculate the distance travelled from t = 4 to t = 5

At $t = 5$:
$s(5) = 4(5)^2 - 0.125(5)^4 = 100 - 0.125(625) = 100 - 78.125 = 21.875\text{ m}$
Distance from $t=4$ to $t=5$ is $|s(5) - s(4)| = |21.875 - 32| = 10.125\text{ m}$

6. Calculate the total distance and average speed

Total distance $= 32 + 10.125 = 42.125\text{ m}$
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{42.125}{5} = 8.425\text{ m/s}$