Tests reveal that a normal driver takes about 0.75 s before he orshecan react to a situation to avoid a collision. it takes about 3 sfora driver having 0.1% alcohol in his system to do the same. ifsuchdrivers are traveling on a straight road at 44 ft/s and theircars can decelerate at 2 ft/s^2, determine the shortest stoppingdistanced for each from the moment they see the pedestrians.

The shortest stopping distance for the normal driver is 517 ft

The shortest stopping distance for the alcoholic driver is 616 ft.

Explanation:

The data, we have is:

Vi = Initial Speed = 44 ft/s

Vf = Final Speed = 0 ft/s (Since, the car finally stops)

a = deceleration = - 2 ft/s²

t = time taken to stop after applying brakes = ?

First, we calculate the time taken by the car to stop after applying the brakes:

Using 1st equation of motion:

Vf = Vi + at

t = (Vf - Vi)/a

t = (0 ft/s - 44 ft/s)/(-2 ft/s²)

t = 22 sec

Now, the stopping distance after applying brakes (S1) is given by 2nd equation of motion:

S1 = Vi t + (1/2)at²

S1 = (44 ft/s)(22 s) + (1/2)(-2 ft/s²)(22 s)²

S1 = 484 ft

FOR NORMAL DRIVER:

Since, the driver takes 0.75 s to respond to a situation, so the distance traveled in this time (S2) is given by:

S2 = Vt

S2 = (44 ft/s)(0.75 s)

S2 = 33 ft

Thus, the total stopping distance for a normal driver is:

S = S1 + S2

S = 484 ft + 33 ft

S = 517 ft

FOR ALCOHOLIC DRIVER:

Since, the driver takes 0.75 s to respond to a situation, so the distance traveled in this time (S2) is given by:

S2 = Vt

S2 = (44 ft/s)(3 s)

S2 = 132 ft

Thus, the total stopping distance for a normal driver is:

S = S1 + S2

S = 484 ft + 132 ft

S = 616 ft

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