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

$$V = \frac{1}{2}(8)^2 - 3(8) + 7$$ $$V = \frac{1}{2}(64) - 24 + 7$$ $$V = 32 - 24 + 7$$ $$V = 8 + 7$$ $$V = 15$$

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

b) The acceleration when $t=0$

$$a = \frac{dV}{dt} = \frac{d}{dt}\left(\frac{1}{2}t^2 - 3t + 7\right)$$ $$a = \frac{1}{2}(2t) - 3 + 0$$ $$a = t - 3$$

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

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

c) The minimum velocity attained

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

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

$$\frac{da}{dt} = \frac{d}{dt}(t - 3) = 1$$

Since $\frac{da}{dt} = 1 > 0$, it is indeed a minimum.
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

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

$$\frac{dy}{dx} = \frac{d}{dx}(x^3 + 3x^2 - 9x - 27)$$ $$\frac{dy}{dx} = 3x^2 + 6x - 9$$

$$3x^2 + 6x - 9 = 0$$

Divide by 3:

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

Factor the quadratic equation:

$$(x+3)(x-1) = 0$$

This gives two possibilities:

$$x+3 = 0 \implies x = -3$$ $$x-1 = 0 \implies x = 1$$

For $x=-3$:

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

So, one stationary point is $(-3,0)$.
For $x=1$:

$$y = (1)^3 + 3(1)^2 - 9(1) - 27$$ $$y = 1 + 3 - 9 - 27$$ $$y = 4 - 36$$ $$y = -32$$

So, the other stationary point is $(1,-32)$.
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 $(-3,0)$ (where $x=-3$):

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

Since $\frac{d^2y}{dx^2} < 0$, the point $(-3,0)$ is a maximum point.
For $(1,-32)$ (where $x=1$):

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

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.
Based on the information:

• The curve passes through x-intercepts $(3,0)$ and $(-3,0)$.
• The curve passes through y-intercept $(0,-27)$.
• There is a maximum point at $(-3,0)$.
• There is a minimum point at $(1,-32)$.

The curve rises from the left, reaches a maximum at $(-3,0)$, then decreases, passes through the y-intercept $(0,-27)$, continues to decrease to the minimum point at $(1,-32)$, and then increases to the right, passing through the x-intercept $(3,0)$. The provided image shows a correct sketch of the curve.