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 .

Answer 1

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 x_t = x₁ + x₂ { eq.3 }

inserting the values in eq.3,

x_t = (4.81 m) + (6.25 m)

x_total = 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 x_t = x₃ + x₄

x_t = (17.5 m) + (82.8 m)

x_total = 100.3 m