How much distance is between you and the deerwhen you come to a stop? How much d

Question

How much distance is between you and the deerwhen you come to a stop? How much distance is between you and the deerwhen you come to a stop? What is the maximum speed you could have andstill not hit the deer? What is the maximum speed you could have andstill not hit the deer? You're driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. Your reaction time before steppingon the brakes is 0.5 s , and the maximum deceleration of your car is 10 m/s^2. How much distance is between you and the deerwhen you come to a stop? What is the maximum speed you could have andstill not hit the deer?

Explanation / Answer

. 01.   The equations related this problem are: 02.      V = (V0) +()*(t)      ;   where = acceleration 03.      D = (D0) +(V0)*(t) + (1/2)*()*(t2) . 04.   The driver's reaction time consumes 1/2seconds; at 20 m/s this consumes 10 meters of roadway. Thus,by the time the brakes activate there is only 25 metersbetween the vehicle and the deer. . 05.   Part a: How much distance isbetween the vehicle and the deer when you come to astop? 06.      V = (V0) +()*(t)   07.      0.0meters/second = (20.0 meters/second) + (-10m/s2)*(t)      ;   solvefor t 08.      t = (-20.0m/s)/(-10m/s2) = 2.0 seconds 09.      D = (D0) +(V0)*(t) + (1/2)*()*(t2) 10.      D = (0.0 meters) + (20.0meters/second)*(2.0 seconds) + (1/2)*(-10 m/s2)*(2.0seconds)2 11.      D = 40.0 meters - 20.0meters = 20.0meters 12.      The vehiclestopped at 20.0 meters from the moment the brakesactivated; the deer is at 25.0 meters from the moment of brakeapplication, thus, there remains 5 meters between the vehicle andthe deer. . 13.   Part b: What maximum speed couldthe vehicle be traveling and still not hit the deer? 14.      V = 0.0 m/s = (V2m/s) + (-10m/s2)*(t2)      ; 15.      D = [35 m -(V2)*(0.5 s)] = (0.0 meters) + (V2meters/second)*(t2 seconds) + (1/2)*(-10m/s2)*(t2 seconds)2 16.      I'm going to drop theunits for clarity, and then bring them back at the end of thesolution. 17.      V = 0.0 = (V2) +(-10)*(t2)           V2 = 10*t2 18.      D = [35 -(V2)*(0.5)] = (V2)*(t2) +(1/2)*(-10)*(t2)2 =(V2)*(t2) +(-5)*(t2)2 19.      Substitue '17' into'18': 20.         [35 -(5*t2)] = (10*t2)*(t2) +(-5)*(t2)2 = (+5)*(t2)2 21.         (+5)*(t2)2 =35 -(5)*t2    ;   writein standard quadratic form 22.         Let = t2 23.        52 + 5 - 35 =0      ; divide through the equationby 5 24.         2+ - 7 =0      ;   solve for using the quadratic formula: { -b ±[b2 - 4ac](1/2) }/(2a) 25.         t= 2.19258 seconds 26.         Using'17' 27.            V2= (10 meters/second2)*(2.19258 seconds) = 21.9258 meters/second . 28.   Let's verify the last result using '03': 29.      D2a = distancebetween the vehicle and the deer at V2 with ahalf-second of reaction time 30.      D2a = [35.0m - (21.9258 m/s)*(0.5s)] = 24.037 meters 31.      D2b = 0.0 m+ (21.9258 m/s)*(2.19258 seconds) + (1/2)*(-10m/s2)*(2.19258 seconds)2 32.      D2b = 0.0m +48.074m - 24.037m = 24.037meters   ;    33.      The exactdistance between the vehicle and deer ! . 16.      I'm going to drop theunits for clarity, and then bring them back at the end of thesolution. 17.      V = 0.0 = (V2) +(-10)*(t2)           V2 = 10*t2 18.      D = [35 -(V2)*(0.5)] = (V2)*(t2) +(1/2)*(-10)*(t2)2 =(V2)*(t2) +(-5)*(t2)2 19.      Substitue '17' into'18': 20.         [35 -(5*t2)] = (10*t2)*(t2) +(-5)*(t2)2 = (+5)*(t2)2 21.         (+5)*(t2)2 =35 -(5)*t2    ;   writein standard quadratic form 22.         Let = t2 23.        52 + 5 - 35 =0      ; divide through the equationby 5 24.         2+ - 7 =0      ;   solve for using the quadratic formula: { -b ±[b2 - 4ac](1/2) }/(2a) 25.         t= 2.19258 seconds 26.         Using'17' 27.            V2= (10 meters/second2)*(2.19258 seconds) = 21.9258 meters/second . 28.   Let's verify the last result using '03': 29.      D2a = distancebetween the vehicle and the deer at V2 with ahalf-second of reaction time 30.      D2a = [35.0m - (21.9258 m/s)*(0.5s)] = 24.037 meters 31.      D2b = 0.0 m+ (21.9258 m/s)*(2.19258 seconds) + (1/2)*(-10m/s2)*(2.19258 seconds)2 32.      D2b = 0.0m +48.074m - 24.037m = 24.037meters   ;    33.      The exactdistance between the vehicle and deer ! 21.         (+5)*(t2)2 =35 -(5)*t2    ;   writein standard quadratic form 22.         Let = t2 23.        52 + 5 - 35 =0      ; divide through the equationby 5 24.         2+ - 7 =0      ;   solve for using the quadratic formula: { -b ±[b2 - 4ac](1/2) }/(2a) 25.         t= 2.19258 seconds 26.         Using'17' 27.            V2= (10 meters/second2)*(2.19258 seconds) = 21.9258 meters/second . 28.   Let's verify the last result using '03': 29.      D2a = distancebetween the vehicle and the deer at V2 with ahalf-second of reaction time 30.      D2a = [35.0m - (21.9258 m/s)*(0.5s)] = 24.037 meters 31.      D2b = 0.0 m+ (21.9258 m/s)*(2.19258 seconds) + (1/2)*(-10m/s2)*(2.19258 seconds)2 32.      D2b = 0.0m +48.074m - 24.037m = 24.037meters   ;    33.      The exactdistance between the vehicle and deer ! .

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