**b.** When $v=2$, we need to impose $u|_{x=2t}=0, t>0$

First when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)

Then when $2t<x<3t$, $x+3t>0$, but $x-3t<0$.

\begin{align}

\phi(x) = \frac{1}{2}\cos(x)+\frac{1}{6}\int_0^x \! \sin(y) dx.\\

\phi(x) = \frac{1}{2}\cos(x) - \frac{1}{6}(\cos(x)-1)\\

\phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}\\

\end{align}

We need to impose the extra condition such that

\begin{align}

u = \phi(5t) + \psi(-t) = 0\\

\phi(5t) = -\psi(-t)\\

\psi(t) = \phi(5t)\\

\psi(x-3t) = \frac{1}{3}\cos(5x-15t) + \frac{1}{6}\\

\end{align}

Then we get $u(x, y) = \frac{1}{3}(\cos(x+3t) +\cos(5x-15t)) + \frac{1}{3}$ for $2t<x<3t$.

**c.** When $v=-4$, we need to impose $u|_{x=-4t}=ux|_{x=-4t}=0, t>0$

Firstly, when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)

Then when $-4t<x<-3t$, $x+3t<0$ and $x-3t<0$.

\begin{align}

\phi(5t) + \psi(-t) = 0\\

\phi'(5t) + \psi'(-t) = 0\\

\end{align}

Then we get $\phi'(5t)=0$ and $\psi(-t) = 0 \implies u(x,t) = 0$ for $-4t<x<-3t$.

Lastly, -3t<x<3t, the situation is similar to part b) $u(x, y) = \frac{1}{3}(\cos(x+3t) +\cos(5x-15t)) + \frac{1}{3}$ .

**d.** When $v=-3$, we need to impose $u|_{x=-3t}=0, t>0$

First, when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)

Then when $ -3t<x<3t$, $x+3t>0$ and $x-3t<0$.

Then $\phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}$

We need to find $\psi(x)$ by imposing the condition:

\begin{align}

\phi(0) + \psi(-6t) = 0\\

0 + \psi(-6t) = 0\\

\psi(x-3t) = 0

\end{align}

Then we get $u(x,t) = \phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}$ for $ -3t<x<3t$.