A particle moves in a straight line so that the velocity is given by V = 1/2t2 - 3t + 7. Find a) the velocity after 8 seconds, b) the acceleration when t=0, c) the minimum velocity attained.

$$V = \frac{1}{2}(64) - 24 + 7$$ $$V = 32 - 24 + 7$$ $$V = 15$$

The velocity after 8 seconds is $\boxed{15 \text{ units/s}}$.

b) The acceleration when $t=0$

$$a = \frac{dV}{dt} = \frac{d}{dt}\left(\frac{1}{2}t^2 - 3t + 7\right)$$ $$a = t - 3$$

$$a = 0 - 3$$ $$a = -3$$

The acceleration when $t=0$ is $\boxed{-3 \text{ units/s}^2}$.

c) The minimum velocity attained

$$a = t - 3 = 0$$ $$t = 3$$

$$V_{min} = \frac{1}{2}(3)^2 - 3(3) + 7$$ $$V_{min} = \frac{1}{2}(9) - 9 + 7$$ $$V_{min} = 4.5 - 9 + 7$$ $$V_{min} = 2.5$$

The minimum velocity attained is $\boxed{2.5 \text{ units/s}}$.

Question 17:
Given the graph $y = (x^2 - 9)(x + 3)$.

a) The x-intercept of the curve

$$0 = (x-3)(x+3)(x+3)$$ $$0 = (x-3)(x+3)^2$$

$$x-3 = 0 \implies x = 3$$ $$x+3 = 0 \implies x = -3$$

The x-intercepts are $\boxed{(3,0) \text{ and } (-3,0)}$.

b) The y-intercept of the curve

$$y = (0^2 - 9)(0 + 3)$$ $$y = (-9)(3)$$ $$y = -27$$

The y-intercept is $\boxed{(0,-27)}$.

c) The stationary point of the curve

$$y = (x^2 - 9)(x + 3)$$ $$y = x^3 + 3x^2 - 9x - 27$$

$$3x^2 + 6x - 9 = 0$$ $$x^2 + 2x - 3 = 0$$ $$(x+3)(x-1) = 0$$ $$x = -3 \text{ or } x = 1$$

For $x=-3$:

$$y = (-3)^3 + 3(-3)^2 - 9(-3) - 27 = -27 + 27 + 27 - 27 = 0$$

For $x=1$:

$$y = (1)^3 + 3(1)^2 - 9(1) - 27 = 1 + 3 - 9 - 27 = -32$$

The stationary points are $\boxed{(-3,0) \text{ and } (1,-32)}$.

d) The nature of the stationary point in c above

$$\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 + 6x - 9)$$ $$\frac{d^2y}{dx^2} = 6x + 6$$

For $x=-3$:

$$\frac{d^2y}{dx^2}\Big|_{x=-3} = 6(-3) + 6 = -18 + 6 = -12$$

Since $\frac{d^2y}{dx^2} < 0$, the point $(-3,0)$ is a maximum point.
For $x=1$:

$$\frac{d^2y}{dx^2}\Big|_{x=1} = 6(1) + 6 = 6 + 6 = 12$$

Since $\frac{d^2y}{dx^2} > 0$, the point $(1,-32)$ is a minimum point.
So, $\boxed{(-3,0) \text{ is a maximum point while } (1,-32) \text{ is a minimum point}}$.

e) Sketch the curve.
The sketch should show:

• X-intercepts at $(3,0)$ and $(-3,0)$.
• Y-intercept at $(0,-27)$.
• A maximum point at $(-3,0)$.
• A minimum point at $(1,-32)$. The curve rises to a maximum at $(-3,0)$, then decreases through $(0,-27)$ to a minimum at $(1,-32)$, and then increases, passing through $(3,0)$. The provided image shows a correct sketch of the curve.