Answer to Question #211833 in Classical Mechanics for eib

Question

A typical automobile under hard braking loses speed at a rate of about 7.0 m/s2 ; the typical reaction time to engage the brakes is 0.50 s. A local school board sets the speed limit in a school zone such that all cars should be able to stop in 4.0 m. (a) What maximum speed does this imply for an automobile in this zone? (b) What fraction of the 4.0 m is due to the reaction time?

Answer 1

$\text{a)let }v_0-\text{initial car speed}$

$v_1=0 -\text{car speed at the end of movement}$

$s = 4 - \text{braking distance}$

$s = s_1+s_2$

$s_1 -\text{distance traveled before driver reaction}$

$s_1= v_0*t_r= 0.5v_0$

$s_2- \text{distance traveled during braking}$

$s_2=\frac{v^2-v_0^2}{2a}=\frac{-v_0^2}{2*(-7)}=\frac{v_0^2}{14}$

$\frac{v_0^2}{14}+0.5v_0=4$

$v_0^2+7v_0-56 =0$

$v_0= 4.76;\text{knock out only a positive value}$

$b) \frac{v_0*t_r}{s}=\frac{4.76*0.5}{4}=0.595=59.5 \%$

$\text{Answer : maximum speed is }4.76\frac{m}{s}$

$\text{fraction of the 4.0 m is due to the reaction time }59.5\%$