Question

Determine the stopping distances for an automobile going a constant initial speed of 93km/h and human reaction time of 0.30 s, for an acceleration a=−3.3m/s2.

Answer 1

From equation of motion

$v_f^2-v_i^2=2a\Delta x\\\ \\\Delta x=\dfrac{v_f^2-v_i^2}{2a}$

Stopping distance $d=vt+\Delta x$

Here, $v_i=93\ km/h=93(5/18)\ m/s= 26\ m/s$

$\Delta x=\dfrac{v_f^2-v_i^2}{2a}=\dfrac{0^2-26^2}{-2\times 3.3}=102.42\ m$

and Stopping Distance $d=vt+Δx =23(0.30)+102.42=109.32\ m$