# Given a known total stopping distance, how can I calculate the initial speed?

This would be my first question on Physics.stackexchange. As I looked closely if I would not double my question, I try dare to ask the question.

I am doing some calculations together with my kids. They had two physics questions on school which I find particularly interesting.

First question: Calculate the total stopping distance for a car knowing the initial speed v in km/h, the friction $\mu$ and the thinking time t of 1 second.

I solved this one pretty easy using the formula: $$d = (v * t) + \dfrac{v^{2}}{2*\mu*g}$$

So given the example that the car is travelling with a speed of 144 km/h, assuming g of 9.81 and a friction of 0.3 this will yield:

$$d = ( 40 * 1) + \dfrac{1600}{2*0.3*9.81} => 311.83146449201496$$

The second question however is where I made up a mistake in my math. This question is, given a known stopping distance d and a friction $\mu$ and still assuming a thinking time of 1 second, what is the initial speed in km/h?

By putting this initial speed and $\mu$ into the first equation of the first question, the same total stopping distance should be proven.

What would be the formula for that last question?

I am assuming that using: $$v = \sqrt{2 * \mu\ * g\ * d }$$

but that is not including the thinking time right? Hence my outcome is not correct.

Well, assuming your first formula is correct, you should be able to use that. We already know the relationship $$d = (v t) + \frac{v^2}{2 \mu g}$$ and if we rearrange slightly we get $$v^2 \frac{1}{2 \mu g} + v(t) - d = 0$$ If you look at that equation you may notice that it is just the quadratic equation for the variable v, meaning it can be solved using the quadratic formula $$x=\frac{-b \pm \sqrt{ b^2 - 4ac}}{2a}$$ where the equation has the format $$0 = ax^2 + bx + c$$(which is the format I rearranged the equation to).
This will give you 2 answers due to the $\pm$ in the equation; but one should be physically insignificant (like a negative velocity). I'll leave you to work out the actual numbers.
You will have to solve the quadratic equation $\frac{1}{2\mu g}v^2 + tv - d = 0$ for which the roots can be found by the quadratic formula.