document.write( "Question 388744: An automobile is travel1ing at 16 m/s. The driver perceives a need to stop and appIies the brakes producing a (negative) acceleration of 4.0 m/S^2. Take the reaction time of the driver, the time to move from the go pedal to the stop pedal, as 1,0 s.
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Algebra.Com's Answer #275123 by richard1234(7193)\"\" \"About 
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Split this into two parts: the time that the car is going at a constant speed, and the time that the car is decelerating.\r
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\n" ); document.write( "\n" ); document.write( "With the driver's reaction time 1 second, the car travels 16 m since it is going at 16 m/s.\r
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\n" ); document.write( "\n" ); document.write( "After, the car decelerates with an acceleration of -4 m/s^2. Since v = at + v_0,\r
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\n" ); document.write( "\n" ); document.write( "\"0+m%2Fs+=+%28-4+%28m%2Fs%5E2%29%29t+%2B+16+m%2Fs\" --> t = 4 seconds. Using the equation\r
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\n" ); document.write( "\n" ); document.write( "\"X+=+%281%2F2%29at%5E2+%2B+%28v_0%29t+%2B+x_0\" we plug in values a = -4 m/s^2, t = 4 s, v_0 = 16 m/s, x_0 = 16 m (since the car already traveled 16 m)\r
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\n" ); document.write( "\n" ); document.write( "\"X+=+%281%2F2%29%28-4+%28m%2Fs%5E2%29%29%284+s%29%5E2+%2B+%2816+m%2Fs%29%284+s%29+%2B+16+m\"\r
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\n" ); document.write( "\n" ); document.write( "\"X+=+48+m\"
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