Question

Determine the stopping distances for an automobile with an initial speed of 95 km/h and human reaction time of 1.0 s : (a) for an acceleration a = -5.5 m/s2 : (b) for a = -6.6 m/s2

Answer 1

(a) 89.8 m

First of all, let's convert the initial speed from km/h into m/s:

$u=95 km/h \cdot \frac{1000 m/km}{3600 s/h}=26.4 m/s$

During the reaction time of $t_1 = 1.0 s$, the car still travels at constant speed, so it covers a distance of

$S_1 = u t_1 = (26.4 m/s)(1.0 s)=26.4 m$

Later, it slows down with acceleration

$a=-5.5 m/s^2$

The distance covered in this second part can be found with the following equation:

$v^2 - u^2 = 2aS_2$

where v=0 is the final speed of the car. Solving for S2,

$S_2 = \frac{v^2-u^2}{2a}=\frac{0-(26.4 m/s)^2}{2(-5.5 m/s^2)}=63.4 m$

So the total stopping distance is

$S=S_1 + S_2 = 26.4 m+63.4 m=89.8 m$

(b) 79.2 m

This exercise is identical to the first part, except that this time the acceleration is

$a=-6.6 m/s^2$

So, the distance covered while decelerating is

$S_2 = \frac{v^2-u^2}{2a}=\frac{0-(26.4 m/s)^2}{2(-6.6 m/s^2)}=52.8 m$

So the total stopping distance is

$S=S_1 + S_2 = 26.4 m+52.8 m=79.2 m$