document.write( "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.
\n" ); document.write( "(a) a = -4.0 m/s^2\r
\n" ); document.write( "\n" ); document.write( "(b) a = -8.0 m/s^2
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Algebra.Com's Answer #707131 by KMST(5328)\"\" \"About 
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IN IDEAS:
\n" ); document.write( "\"%2293+km+%2F+h%22=93%2A1000%2F3600\"\"%22m+%2F+s%22=93%2F3.6\"\"%22m+%2F+s+%22=%22about+25.833+m+%2F+s%22\" .
\n" ); document.write( "At that speed, during the 3.0 seconds allowed for reaction time,
\n" ); document.write( "the vehicle would cover a distance of
\n" ); document.write( "\"93%2F3.6\"\"%22m+%2F+s%22%2A%223.0+s%22=77.5m\" .\r
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\n" ); document.write( "\n" ); document.write( "The graph of speed (in m/s) as a function of time (in seconds)
\n" ); document.write( "for the first case (a=-4.0m/s^2) is
\n" ); document.write( " ,
\n" ); document.write( "and the stopping distance is the area under the curve.
\n" ); document.write( "To go from initial speedspeed to \"%220+m+%2F+s%22\"
\n" ); document.write( "at \"a=%22-4.0%22\"\"m%2Fs%5E2\" it would take \"%28%2893%2F3.6%29%29%2F4.0\"\"s=93%2F14.4\"\"s\" .
\n" ); document.write( "During that time, speed would be changing linearly,
\n" ); document.write( "with an average speed of \"%2893%2F3.6%2B0%29%2F2\"\"%22m+%2F+s%22=93%2F7.2\"\"%22m+%2F+s%22=%22about12.917+m+%2F+s%22\" ,
\n" ); document.write( "and covering a distance of
\n" ); document.write( "\"%2893%2F14.4%29%2A%2893%2F7.2%29\"\"m=%22about+83.42+m%22\" .
\n" ); document.write( "The total stopping distance would be about
\n" ); document.write( "\"77.5m%2B83.42m=160.92m\"
\n" ); document.write( "As the acceleration and initial speed were given with two significant digits,
\n" ); document.write( "it would be proper to report the stopping distance as
\n" ); document.write( "\"highlight%28161%29m\" .
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\n" ); document.write( "With an acceleration of -8.0 m/s^2 (twice as large in magnitude),
\n" ); document.write( "the stopping time would be \"93%2F38.8\"\"s=%22about+3.229+s%22\" .
\n" ); document.write( "During that time, with the same average speed calculated above,
\n" ); document.write( "the vehicle would cover a distance of
\n" ); document.write( "\"%2893%2F14.4%29%2A%2893%2F38.8%29\"\"m=%22about+41.71+m%22\" .
\n" ); document.write( "That results in a total stooping distance of
\n" ); document.write( "\"77.5m%2B41.72m=119.2m\" to be reported as \"highlight%28119m%29\" .
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\n" ); document.write( "IN FORMULAS:
\n" ); document.write( "With d=distance, v=velocity, a=acceleration, and t=time, in SI units (only meters and seconds allowed),
\n" ); document.write( "\"d=v%2At\" for uniform linear motion (constant speed for the 3.0 seconds of reaction time), and
\n" ); document.write( "\"d=v%2At%2Ba%2At%5E2%2F2\" for uniformly accelerated linear motion (the braking part).
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