Let's solve the problem step by step.

Part (a): Determine constants $a$, $b$, and $c$

Given:

• $y(t) = -\frac{1}{3}(t - 3) + 2$ for $t$ between 0 and 6.
• The signal $z(t) = ax(bt + c)$.

We need to determine the constants $a$, $b$, and $c$ so that the signal $z(t)$ matches the form in the figure.

The figure for $z(t)$ is a piecewise function composed of linear segments:

1. For $t$ between $-4$ and $0$, the line has a positive slope.
2. For $t$ between $0$ and $6$, the line is flat at $3$.

Let’s assume $z(t) = ax(bt + c)$ can be written as:

$z(t) = at + b$

To match the given $z(t)$ with the shape in the figure:

• At $t = 0$, $z(t) = 3$, hence: $3 = a(0) + b \Rightarrow b = 3$

Now, consider the line segment from $t = -4$ to $t = 0$:

• For $t = -4$, the value of $z(t)$ is $0$, so: $0 = a(-4) + 3 \Rightarrow -4a = -3 \Rightarrow a = \frac{3}{4}$

Therefore, $z(t)$ can be expressed as:

$z(t) = \frac{3}{4}t + 3$

Part (b): Determine and sketch a signal $v(t)$ such that $z(t) = \int_{-\infty}^{t} v(\tau) d\tau$

Since $z(t)$ is the integral of $v(t)$, we need to find the derivative of $z(t)$ to obtain $v(t)$:

1. For $t$ in the interval $[-4, 0]$: $z(t) = \frac{3}{4}t + 3$ Therefore, $v(t) = \frac{dz(t)}{dt} = \frac{3}{4}$

2. For $t$ in the interval $[0, 6]$: $z(t) = 3$ Therefore, $v(t) = \frac{dz(t)}{dt} = 0$

Thus, the signal $v(t)$ can be represented as: