A particle moves in the $xy$-plane so that $x=at$ and $y=at(1-bt)$, where $a, b>0$.
(a) Find the path $y(x)$.
(b) Find the velocity $\textbf{v}$ and acceleration $\textbf{a}$ and their magnitudes as functions of time.
(c) Find the time $t_0$ at which the angle between $\textbf{v}$ and $\textbf{a}$ is $45°$.
I just don't understand how to do this. I've given it a go and got $\textbf{v}=a \textbf{i}+a-2abt\textbf{j}$ and $\textbf{a}=-2ab\textbf{j}$. Then I can't figure out the magnitudes of them either. Any help would be appreciated.
For (a), using the first equation to solve for $t$ gives $t=\frac{x}{a}$. Plugging this into the second expression gives $$y(x)=a\frac{x}{a}(1-b\frac{x}{a})=x(1-\frac{b}{a}x)$$ For (b), to find $\textbf{v}$, take the derivatives of both components with respect to time (as you've done) to get $\textbf{v}=\langle a, a-2abt \rangle$. Thus $\textbf{v}$ has a magnitude of $\sqrt{a^2+(a-2abt)^2}$. To find $\textbf{a}$, take the derivatives of both components of $\textbf{v}$ to get $\textbf{a}=\langle 0,-2ab\rangle$. Thus $\textbf{a}$ has magnitude $\sqrt{0^2+(-2ab)^2}=2ab$.
For (c), if the angle $\theta$ between the vectors is $45°$, then $\cos{\theta}=\frac{\sqrt{2}}{2}$. Now, you can use the dot product since $\textbf{v} \cdot \textbf{a}=||\textbf{v}|| \,||\textbf{a}||\cos{\theta}$. Replacing everything in this equation with our found quantities gives $\langle a,a-2abt \rangle \cdot \langle 0, -2ab \rangle=ab\sqrt{2a^2+2(a-2abt)^2}$. This can be rewritten as $$-2ab(a-2abt)=ab\sqrt{2a^2+2(a-2abt)^2}$$Solving this expression for $t$ gives you the time at which the angle between $\textbf{v}$ and $\textbf{a}$ is $45°$.