A scooter travelling up and down a straight horizontal road passes a fixed point O with a speed of $p\ \text{m s}^{-1}$. The velocity, $v\ \text{m s}^{-1}$, of the scooter, $t$ seconds after passing O, is given by $v = 7 - 8\sin \tfrac{1}{3}t.$

1. Substitute $t = 0$ into the velocity equation to find the initial speed

At point O, $t = 0$.
$v = 7 - 8\sin(0)$
$v = 7 - 0 = 7$

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

$v = 0$
$7 - 8\sin \left(\frac{1}{3}t\right) = 0$
$\sin \left(\frac{1}{3}t\right) = \frac{7}{8}$

2. Find the basic angle in radians

$\text{Basic angle} = \sin^{-1}\left(\frac{7}{8}\right) \approx 1.0654 \text{ rad}$

3. Find the first value of $t$

$\frac{1}{3}t = 1.0654$
$t = 3.1962 \approx 3.20$

4. Find the second value of $t$

$\frac{1}{3}t = \pi - 1.0654$
$\frac{1}{3}t = 2.0762$
$t = 6.2286 \approx 6.23$

1. Identify the time interval for the third second and verify the sign of velocity.

The third second corresponds to the time interval from $t = 2$ to $t = 3$.
From part (ii), the first time the scooter changes direction ($v = 0$) is at $t = 3\sin^{-1}(\frac{7}{8}) \approx 3.20\text{ s}$.
Since $3.20 > 3$, the velocity does not change sign and remains positive during the interval $2 \leqslant t \leqslant 3$.

2. Set up the definite integral for the distance travelled.

$\text{Distance} = \int_{2}^{3} \left(7 - 8\sin \tfrac{1}{3}t\right) \, dt$

3. Integrate the velocity function.

$\int \left(7 - 8\sin \tfrac{1}{3}t\right) \, dt = 7t - \frac{8(-\cos \tfrac{1}{3}t)}{\frac{1}{3}} = 7t + 24\cos \tfrac{1}{3}t$

4. Evaluate the definite integral.

$\text{Distance} = \left[ 7t + 24\cos \tfrac{1}{3}t \right]_{2}^{3}$
$= \left( 7(3) + 24\cos 1 \right) - \left( 7(2) + 24\cos \tfrac{2}{3} \right)$
$= (21 + 12.967) - (14 + 18.861)$
$= 33.967 - 32.861 = 1.106 \approx 1.11\text{ m}$