The problem provided asks to solve for velocity and acceleration based on the displacement function:

Given displacement function:

$s(t) = 4t^3 - 18t^2 + 5t + 10$

where $s$ is the displacement of the particle in meters at time $t$ seconds.

Part (a): Find expressions for velocity $v$ and acceleration $a$.

(i) Velocity $v(t)$:

Velocity is the rate of change of displacement with respect to time, which is the first derivative of $s(t)$.

$v(t) = \frac{ds}{dt} = \frac{d}{dt}(4t^3 - 18t^2 + 5t + 10)$ Let's differentiate:

$v(t) = 12t^2 - 36t + 5$

So, the velocity expression is: $v(t) = 12t^2 - 36t + 5$

(ii) Acceleration $a(t)$:

Acceleration is the rate of change of velocity with respect to time, which is the second derivative of $s(t)$ or the first derivative of $v(t)$.

$a(t) = \frac{dv}{dt} = \frac{d}{dt}(12t^2 - 36t + 5)$ Let's differentiate:

$a(t) = 24t - 36$

So, the acceleration expression is: $a(t) = 24t - 36$

Part (b): Find the range of values of $t$ for which the acceleration is positive.

For acceleration to be positive: $a(t) > 0$ Substitute $a(t) = 24t - 36$:

$24t - 36 > 0$ Solving for $t$:

$24t > 36 \quad \Rightarrow \quad t > \frac{36}{24} = 1.5$

Thus, the range of $t$ for which the acceleration is positive is: $t > 1.5$

Would you like a more detailed explanation on any step? Here are five related questions for further practice:

1. What is the displacement when $t = 2$ seconds?
2. At what time does the velocity become zero?
3. How would the graph of acceleration $a(t) = 24t - 36$ look?
4. What is the maximum or minimum value of displacement?
5. Find the total distance traveled by the particle in the first 4 seconds.

Tip: Remember, velocity is the slope of the displacement curve, while acceleration is the slope of the velocity curve!