Question 1092443
IN IDEAS:
{{{"93 km / h"=93*1000/3600}}}{{{"m / s"=93/3.6}}}{{{"m / s "="about 25.833 m / s"}}} .
At that speed, during the 3.0 seconds allowed for reaction time,
the vehicle would cover a distance of
{{{93/3.6}}}{{{"m / s"*"3.0 s"=77.5m}}} .


The graph of speed (in m/s) as a function of time (in seconds)
for the first case (a=-4.0m/s^2) is
{{{drawing(600,300,-2,8,-3,27,grid(0),
line(0,25.83,3,25.83),line(3,25.83,6.458,0)
)}}} ,
and the stopping distance is the area under the curve.
To go from initial speedspeed to {{{"0 m / s"}}}
at {{{a="-4.0"}}}{{{m/s^2}}} it would take {{{((93/3.6))/4.0}}}{{{s=93/14.4}}}{{{s}}} .
During that time, speed would be changing linearly,
with an average speed of {{{(93/3.6+0)/2}}}{{{"m / s"=93/7.2}}}{{{"m / s"="about12.917 m / s"}}} ,
and covering a distance of
{{{(93/14.4)*(93/7.2)}}}{{{m="about 83.42 m"}}} .
The total stopping distance would be about
{{{77.5m+83.42m=160.92m}}}
As the acceleration and initial speed were given with two significant digits,
it would be proper to report the stopping distance as
{{{highlight(161)m}}} .
 
With an acceleration of -8.0 m/s^2 (twice as large in magnitude),
the stopping time would be {{{93/38.8}}}{{{s="about 3.229 s"}}} .
During that time, with the same average speed calculated above,
the vehicle would cover a distance of
{{{(93/14.4)*(93/38.8)}}}{{{m="about 41.71 m"}}} .
That results in a total stooping distance of
{{{77.5m+41.72m=119.2m}}} to be reported as {{{highlight(119m)}}} .
 
IN FORMULAS:
With d=distance, v=velocity, a=acceleration, and t=time, in SI units (only meters and seconds allowed),
{{{d=v*t}}} for uniform linear motion (constant speed for the 3.0 seconds of reaction time), and
{{{d=v*t+a*t^2/2}}} for uniformly accelerated linear motion (the braking part).