Saturday

December 20, 2014

December 20, 2014

Posted by **Sarah** on Friday, October 11, 2013 at 2:36am.

Determine the stopping distances for an automobile with an initial speed of 89 km/h and human reaction time of 2.0 s for the following:

(a) an acceleration a = -4.0 m/s2.

(b) a = -8.0 m/s2.

- Physics -
**Henry**, Saturday, October 12, 2013 at 10:12pmVo=85000m/h = 85000m/3600s = 23.61 m/s.

a. V = Vo + a*t

T = (V-Vo)/a + Tr

T = (0-23.61)/-4 + 2s = 7.9 s = Stopping time.

d = Vo*t + 0.5a*t^2

d = 23.61*7.9 + 0.5*-4*7.9^2 = 61.7m

b. T = (0-23.61)/-8 + 2s = 4.95 s.

d=23.61*4.95 + 0.5*(-8)*4.95^2=18.86 m.

- Physics -
**Henry**, Saturday, October 12, 2013 at 10:44pmCORRECTION:

Vo=89,000m/1h = 89000/3600s = 24.72 m/s

a. V = Vo + va*t

T = (V-V0)/a + Tr

T = (0-24.72)/-4 + 2s.=8.18 s.=Stopping

time.

d = Vo*t + 0.5a*t^2

d=24.72*8.18 + 0.5*(-40*8.18*2=68.38 m.

b. T = (0-24.72)/-8 + 2s = 5.09 s.

d=24.72*5.09 + 0.5*(-8)*5.09^2= 22.19 m.

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