Tests reveal that a normal driver takes about 0.75 s before he or she can react

Question

Tests reveal that a normal driver takes about 0.75 s before he or she can react to a situation to avoid a collision. It takes about 3 s for a driver having 0.1% alcohol in his system to do the same.(Figure 1) Part A If such drivers are travelling on a straight road at 30 mph (44 ft/s) and their cars can decelerate at 21 ft/s2. determine the shortest stopping distance d for normal driver from the moment he or she see the pedestrians. Part B Determine the shortest stopping distance d for drunk driver from the moment he or she see the pedestrians. Moral: If you must drink, please don't drive!

Explanation / Answer

Given:

Part A

The reaction time t =.75s

v= 44 ft/s

ditance covered during reaction Dr = .75 * 44 = 33 ft

the remaining distance covered as deceleration with a =21 ft/s2

2aS=v2-u2

where final velocity v will be zero u =44 ft/s

therefore S = -442/2* 21 =46.095 ft

therefore the total shortest distance will be 46.095 ft + 33 ft = 79.095 ft

Part B

t= 3 sec

now similarly

ditance covered during reaction Dr = 3 * 44 = 132 ft

again Similarly

the remaining distance covered as deceleration with a =21 ft/s2

2aS=v2-u2 = 46.095 ft (same as above)

Total shortest distance will be 132 + 46.095= 178.095 ft Answer

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