A particle's acceleration is given by a = (6t − 18) m/s2, and its initial velocity is V0 = 24 m/s (at t=0). Find a) the velocity V of the particle at any time t, and b) the values of t at maximum disp

Velocity $V(t)$ is the integral of acceleration $a(t)$ with respect to time $t$.

Given that the initial velocity is $24 \text{ m/s}$ at $t=0$:

The velocity of the particle at any time $t$ is $\boxed{V(t) = 3t^2 - 18t + 24 \text{ m/s}}$.

Set $V(t) = 0$:

Divide the entire equation by 3:

Factor the quadratic equation:

This gives two possible values for $t$:

The values of $t$ at maximum displacement are $\boxed{t=2 \text{ s and } t=4 \text{ s}}$.

For calculating distance, the constant $D_0$ is not needed as it cancels out in differences.

From part (b), we know $V(t)=0$ at $t=2$ s and $t=4$ s. These are the points where the particle changes direction.
We need to calculate the distance in the intervals $[1, 2]$, $[2, 4]$, and $[4, 5]$.

• For the interval $[1, 2]$:

Distance for $[1, 2]$ is $|4| = 4 \text{ m}$.

• For the interval $[2, 4]$:

Distance for $[2, 4]$ is $|-4| = 4 \text{ m}$.

• For the interval $[4, 5]$:

Distance for $[4, 5]$ is $|4| = 4 \text{ m}$.

Total distance $= 4 \text{ m} + 4 \text{ m} + 4 \text{ m} = 12 \text{ m}$.
The distance covered by the particle between $t=1$ second and $t=5$ seconds is $\boxed{12 \text{ m}}$.