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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.

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2 Answers 2

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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.

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    $\begingroup$ Thank you! Solved it, how silly that I could miss this... I am getting way too rusty... $\endgroup$
    – RvdV79
    Commented Feb 3, 2017 at 19:16
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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.

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