A particle moves in a straight line with a fixed point $O$. The velocity of the particle, $v$ m/s, is given by $v = \dfrac{25}{2t+1} + 10$ where $t$ is time in seconds after passing $O$.

1. Explain why the particle never changes its direction of motion

For $t \ge 0$, $2t+1 > 0$, which means $\frac{25}{2t+1} > 0$.
Therefore, $v = \frac{25}{2t+1} + 10 > 10 > 0$ for all $t \ge 0$.
Since the velocity $v$ is always positive, the particle never changes its direction of motion.

2. Explain why the speed eventually approaches 10 m/s

As time $t$ becomes very large ($t \to \infty$), the denominator $2t+1$ becomes very large, so the term $\frac{25}{2t+1}$ approaches $0$.
Thus, $v \to 0 + 10 = 10$.
Therefore, the particle will eventually travel at a speed close to $10$ m/s.

1. Differentiate the velocity function to find the acceleration function

$v = 25(2t+1)^{-1} + 10$

$a = \frac{dv}{dt} = 25(-1)(2t+1)^{-2}(2)$

$a = -50(2t+1)^{-2} = -\frac{50}{(2t+1)^2}$

2. Evaluate the acceleration at $t = 2$

When $t = 2$,
$a = -\frac{50}{(2(2)+1)^2}$
$a = -\frac{50}{5^2}$
$a = -\frac{50}{25} = -2$

1. Set up the definite integral for the distance travelled in the fifth second.

The fifth second is the time interval from $t = 4$ to $t = 5$. Since $v > 0$ for all $t \ge 0$, the distance is given by:
$\text{Distance} = \int_{4}^{5} \left( \frac{25}{2t+1} + 10 \right) dt$

2. Integrate the velocity function.

$\int \left( \frac{25}{2t+1} + 10 \right) dt = \frac{25}{2} \ln(2t+1) + 10t$

3. Evaluate the definite integral using the limits.

$\text{Distance} = \left[ \frac{25}{2} \ln(2t+1) + 10t \right]_{4}^{5}$
$= \left( \frac{25}{2} \ln(11) + 50 \right) - \left( \frac{25}{2} \ln(9) + 40 \right)$
$= \frac{25}{2} (\ln 11 - \ln 9) + 10$
$= 12.5 \ln\left(\frac{11}{9}\right) + 10$

4. Calculate the final numerical value.

$\text{Distance} \approx 12.508 \approx 12.5 \text{ m (to 3 s.f.)}$