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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!

A particle P is moving along a straight line. At time t seconds, the displacement, s meters, of P from a fixed point O on the line is given by s = 4t^3 - 18t^2 + 5t + 10. (a) Find an expression for (i) velocity and (ii) acceleration. (b) Find the range of values of t for which the acceleration of P is positive.

This problem involves solving for the velocity and acceleration of a particle moving along a straight line using calculus. We find the expressions for velocity and acceleration from the displacement function s(t) = 4t^3 - 18t^2 + 5t + 10, and determine the range of t for positive acceleration.

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