A particle $P$ travels in a straight line and its velocity, $v$ m/s, is given by
$v = 4\sin^2\left(\frac{t}{2}\right) - 1$, where $t$ is the time in seconds after passing a fixed point O. Find

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

When $t = 0$,
$v = 4\sin^2\left(\frac{0}{2}\right) - 1$

2. Evaluate the expression

$v = 4(0) - 1 = -1$

1. Find the expression for acceleration by differentiating velocity with respect to time.

$a = \frac{dv}{dt}$
$a = 4 \left[ 2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right) \cdot \frac{1}{2} \right]$
$a = 4\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)$

2. Simplify the acceleration expression using the double angle formula.

$a = 2 \left[ 2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right) \right]$
$a = 2\sin(t)$

3. Set acceleration to 2 and solve for the first time $t$.

$2\sin(t) = 2$
$\sin(t) = 1$
$t = \frac{\pi}{2}$

4. Substitute the value of $t$ back into the velocity equation.

$v = 4\sin^2\left(\frac{\pi/2}{2}\right) - 1$
$v = 4\sin^2\left(\frac{\pi}{4}\right) - 1$
$v = 4\left(\frac{1}{\sqrt{2}}\right)^2 - 1$
$v = 4\left(\frac{1}{2}\right) - 1$
$v = 2 - 1 = 1$

1. Set velocity to zero to find when the particle is at rest

$v = 4\sin^2\left(\frac{t}{2}\right) - 1 = 0$

2. Solve for the trigonometric term

$\sin^2\left(\frac{t}{2}\right) = \frac{1}{4}$
$\sin\left(\frac{t}{2}\right) = \pm \frac{1}{2}$

3. Find the first positive value of t

For the first time after passing O, we take the smallest positive value:
$\sin\left(\frac{t}{2}\right) = \frac{1}{2}$
$\frac{t}{2} = \frac{\pi}{6}$
$t = \frac{\pi}{3}$

1. Express velocity in an integrable form using double angle formula

$v = 4\sin^2\left(\frac{t}{2}\right) - 1$
Using the double angle formula $\cos t = 1 - 2\sin^2\left(\frac{t}{2}\right)$, we have $2\sin^2\left(\frac{t}{2}\right) = 1 - \cos t$.
$v = 2(1 - \cos t) - 1 = 1 - 2\cos t$

2. Check for changes in direction during the 3rd second

The $3^{\text{rd}}$ second is the time interval from $t = 2$ to $t = 3$.
We check if $v = 0$ in this interval:
$1 - 2\cos t = 0 \implies \cos t = 0.5$
$t = \frac{\pi}{3} \approx 1.047 \text{ s} \quad \text{or} \quad t = \frac{5\pi}{3} \approx 5.236 \text{ s}$
Since there are no roots between $t = 2$ and $t = 3$, the particle does not change direction during the $3^{\text{rd}}$ second. Furthermore, $v(2) = 1 - 2\cos 2 > 0$, so the velocity is positive throughout.

3. Integrate velocity to find the total distance travelled

$\text{Distance} = \int_2^3 (1 - 2\cos t) \, dt$
$= \left[ t - 2\sin t \right]_2^3$
$= (3 - 2\sin 3) - (2 - 2\sin 2)$
$= 1 - 2\sin 3 + 2\sin 2$
$= 1 - 2(0.14112) + 2(0.90930)$
$= 2.53636 \dots \approx 2.54 \text{ m}$