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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.

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hamzaahmeds

Answer:

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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