We are working with the displacement function of the particle given as $s(t) = 3\sin(\pi t) + 3\cos(\pi t)$, and we need to find the average velocity over specific time intervals and estimate the instantaneous velocity at $t = 1$.

(a) Average velocity:

The average velocity over a time interval $[t_1, t_2]$ is given by the formula: $v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$ We already know your answers for parts (i) through (iii) are correct, so we'll focus on part (iv), and then proceed to part (b).

(iv) Interval $[1, 1.001]$

Let's calculate the average velocity for this interval:

• $t_1 = 1$
• $t_2 = 1.001$

First, we need to calculate $s(1)$ and $s(1.001)$.

$s(1) = 3\sin(\pi \cdot 1) + 3\cos(\pi \cdot 1) = 3\sin(\pi) + 3\cos(\pi) = 3(0) + 3(-1) = -3 \, \text{cm}$

Now calculate $s(1.001)$: $s(1.001) = 3\sin(\pi \cdot 1.001) + 3\cos(\pi \cdot 1.001)$ We need to approximate $\sin(\pi \cdot 1.001)$ and $\cos(\pi \cdot 1.001)$: $\sin(\pi \cdot 1.001) \approx -0.00314, \quad \cos(\pi \cdot 1.001) \approx -0.999995$ $s(1.001) = 3(-0.00314) + 3(-0.999995) = -0.00942 - 2.999985 = -3.009405 \, \text{cm}$

Now, apply the average velocity formula: $v_{\text{avg}} = \frac{s(1.001) - s(1)}{1.001 - 1} = \frac{-3.009405 - (-3)}{0.001} = \frac{-3.009405 + 3}{0.001} = \frac{-0.009405}{0.001} = -9.41 \, \text{cm/s}$

So, the correct average velocity for this interval is $v_{\text{avg}} = -9.41 \, \text{cm/s}$.

(b) Instantaneous velocity at $t = 1$

The instantaneous velocity is the derivative of the displacement function $s(t)$ with respect to time $t$, i.e., $v(t) = s'(t)$.

The displacement function is: $s(t) = 3\sin(\pi t) + 3\cos(\pi t)$ The derivative $v(t)$ is: $v(t) = \frac{d}{dt} [3\sin(\pi t) + 3\cos(\pi t)] = 3\pi \cos(\pi t) - 3\pi \sin(\pi t)$ At $t = 1$: $v(1) = 3\pi \cos(\pi \cdot 1) - 3\pi \sin(\pi \cdot 1) = 3\pi (-1) - 3\pi (0) = -3\pi$ Approximating $\pi \approx 3.14159$: