I'm having trouble properly conceptualizing a question.

The question is:

"Suppose you are driving down a steep hill (5 degrees, 8.7%, which is fairly steep for roads), on a rainy day ($\mu$ = .5) at a reckless 30 $\frac{m}{s}$. The stoplight at the bottom of the high turns red. It is 75 meters in front of you. (You can neglect reaction time, because you saw it turn yellow, so you were ready to put on breaks.) Can you stop before the light? If so, what is your stopping distance?"

So far, I was able to find out the acceleration in the x distance.

$\sum{}^{}F_x = F_b-F_f \\ ma_x = mg\sin5 - .5mg\cos5 $

The masses cancel out so we are left with:

$a_x = (9.8)\sin5 - .5(9.8)\cos5 = -4.03 \frac{m}{s^2} $

I have no idea where to continue from here. Any help is appreciated. Thanks.

Also I attached a photo: enter image description here


1 Answer 1


The trick to this question is to realize there is no need for a complicated dynamic rules. You took your time to evaluate the friction force acting on a car

$$F = mg \sin(\alpha) - \mu mg \cos(\alpha) $$

and you can see that this force is contants, therefore your acceleration $a = F/m = g( \sin(\alpha) - \mu \cos(\alpha))$ is also constant. Now you can work with uniformly accelerated motion. You know (for $t_f$ being final time)

$$s = v_0t_f + \frac 12 a t_f^2, \qquad 0 = v_0 + at_f$$

which leads to the time of breaking $t_f = -v_0/a$ and therefore the distance of breaking

$$ s = -\frac{v_0^2}a + \frac 12 a \left( \frac{v_0}a \right)^2 = -\frac{v_0^2}{2a}. $$

You can see that the minus is fine, because acceleration is negative in this notation.

  • 1
    $\begingroup$ you dont have to go though ccalculating the time - simply use the equation for uniform accel. as in @dr jh answer $s=u^2/2a$. $\endgroup$
    – user276350
    Oct 20, 2020 at 0:23
  • $\begingroup$ Except the formulas I am presenting are the basic formulas and $s = u^2/2a$ is a derived formula from those two I have presented. There are other methods how to calculate this exercise, but none good of them is to "remember derived formulas". There is no understanding in writing this down. And he does not know how to calculate it, so he has to learn it. And the last problem with your formula is, that it is simply wrong for this case. You will get negative distance. That's because deriving $s = u^2/2a$ assumes something not valid here. $\endgroup$ Oct 20, 2020 at 7:59
  • $\begingroup$ the equation is not wrong. you will not get an incorrect sign since $0 = u^2 +2as$ so $u^2 = - 2as$ and because $a<0$ you will get a positive displacement. Completely valid. $\endgroup$
    – user276724
    Oct 20, 2020 at 23:48
  • $\begingroup$ Yes, the original formula is valid, but not what @max_fisics proposed. $\endgroup$ Oct 24, 2020 at 7:28

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