We were given a couple formulas, but the one that immediately stood out to me was the Vfinal = Vinitial + at
so we know the patrol will constantly accelerate "until he pulls next to the speeding car" so Vfinal = 30m/s and Vinitial = 0(cop at rest) and the acceleration is a constant 3m/s^2
so 30 = 3t, t = 10. However when I continue through this lecture, it turns out t = 20. he uses the formula x = xo + v0t + 1/2at^2 and somehow gets a different time than me. I don't understand how it's possible that we are both solving for time but getting different results.
Is it safe to assume that when xo = 0 and v0 = 0 that
sqrt(2x/a) = Vfinal / a
since we cancel out terms and they both equal t?
Your mistake was that you tried finding the time it takes for the cop to achieve the same velocity ($30 \frac{m}{s}$). That’s not right for obvious reasons. The speeding car was already traveling at $30 \frac{m}{s}$ by the time the police car was stationary. Instead, you have to find the time it takes for the distance covered to become equal.
Car $1$: $\overline v = 30 \frac{m}{s} \implies \Delta x_1 = \overline v \cdot \Delta t$
Car $2$: $v_1 = 0 \frac{m}{s}; \overline a = 3 \frac{m}{s^2} \implies \Delta x_2 = \frac{1}{2}\overline a\Delta t^2+v_1\Delta t$
Keep in mind $\Delta x$ represents displacement, $\Delta t$ represents time, $v_1$ represents initial velocity, $\overline v$ represents average/uniform velocity, and $\overline a$ represents average/uniform acceleration.
We need to find the $\Delta t$ for which $\Delta x_1 = \Delta x_2$.
$$\Delta x_1 = \Delta x_2$$ $$\overline v\cdot\Delta t= \frac{1}{2}\overline a\Delta t^2+v_1\Delta t$$ $$30\Delta t = \frac{1}{2}3\Delta t^2+0(\Delta t)$$ $$30\Delta t = \frac{1}{2}3\Delta t^2$$ $$30 = 1.5\Delta t$$ $$\boxed{\Delta t = \frac{30}{1.5} = 20s}$$