The \"reaction time\" of the average automobile driver is about 0.7 s . (The rea

Question

The "reaction time" of the average automobile driver is about 0.7 s . (The reaction time is the interval between the perception of a signal to stop and the application of the brakes.) An automobile can slow down with an acceleration of 12.4 ft/s2 .

Part A

Compute the total distance covered in coming to a stop after a signal is observed from an initial velocity of 15.4 mi/h . (in a school zone)

Part B

Compute the total distance covered in coming to a stop after a signal is observed from an initial velocity of 56.0 mi/h .

Explanation / Answer

Given that :

The "reaction time" of the average automobile driver, t = 0.7 sec

acceleration, a = 12.4 ft/s2 = 3.78 m/s2

Part-A : the total distance covered in coming to a stop after a signal is observed from an initial velocity which will be given as -

distance covered in reaction time, x1 = v0 t                                                           { eq.1 }

where, v0 = initial velocity = 15.4 mi/hr = 6.88 m/s

inserting the values in above eq.

x1 = (6.88 m/s) (0.7 s)

x1 = 4.81 m

distance covered after applying brakes to stop is given as,

using equation of motion 3, v2 = v02 - 2 a x2 { eq.2 }

where, v = final velocity = 0 m/s

inserting the values in eq.2,

(0 m/s)2 = (6.88 m/s)2 - 2 (3.78 m/s2) x2

47.3 m2/s2 = (7.56 m/s2) x2

x2 = (47.3 m2/s2) / (7.56 m/s2)

x2 = 6.25 m

Now, Total distance covered for stopping xt = x1 + x2 { eq.3 }

inserting the values in eq.3,

xt = (4.81 m) + (6.25 m)

xtotal = 11.06 m

Part-B : the total distance covered in coming to a stop after a signal is observed from an initial velocity which will be given as -

same procedure of part-A which use in part-B.

distance covered in reaction time, x3 = v0 t                                                    

where, v0 = initial velocity = 56 mi/hr = 25.03 m/s

x3 = (25.03 m/s) (0.7 s)

x3 = 17.5 m

distance covered after applying brakes to stop is given as,

using equation of motion 3, v2 = v02 - 2 a x4   

where, v = final velocity = 0 m/s

(0 m/s)2 = (25.03 m/s)2 - 2 (3.78 m/s2) x4   

626.5 m2/s2 = (7.56 m/s2) x4

x4= (626.5 m2/s2) / (7.56 m/s2)

x4 = 82.8 m

Now, Total distance covered for stopping xt = x3 + x4

xt = (17.5 m) + (82.8 m)

xtotal = 100.3 m

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