SOLUTION: Determine the stopping distances for a car with an initial speed of 93 km/h and human reaction time of 3.0 s for the following accelerations.
(a) a = -4.0 m/s^2
(b) a = -8.0 m/
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Question 1092443: Determine the stopping distances for a car with an initial speed of 93 km/h and human reaction time of 3.0 s for the following accelerations.
(a) a = -4.0 m/s^2
(b) a = -8.0 m/s^2
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
IN IDEAS:
.
At that speed, during the 3.0 seconds allowed for reaction time,
the vehicle would cover a distance of
.
The graph of speed (in m/s) as a function of time (in seconds)
for the first case (a=-4.0m/s^2) is
,
and the stopping distance is the area under the curve.
To go from initial speedspeed to
at it would take .
During that time, speed would be changing linearly,
with an average speed of ,
and covering a distance of
.
The total stopping distance would be about
As the acceleration and initial speed were given with two significant digits,
it would be proper to report the stopping distance as
.
With an acceleration of -8.0 m/s^2 (twice as large in magnitude),
the stopping time would be .
During that time, with the same average speed calculated above,
the vehicle would cover a distance of
.
That results in a total stooping distance of
to be reported as .
IN FORMULAS:
With d=distance, v=velocity, a=acceleration, and t=time, in SI units (only meters and seconds allowed),
for uniform linear motion (constant speed for the 3.0 seconds of reaction time), and
for uniformly accelerated linear motion (the braking part).
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