A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft behind the truck. The driver of the car wants to pass the truck, i. e., he wishes to place his car at B, 40 ft in front of the truck, and then resume the speed of 35 mi/h; The maximum acceleration of the car is 5 ft/s2 and the maximum deceleration obtained by applying the brakes is 20 ft/s2. What is the shortest time in which the driver of the car can complete the passing operation? (assuming that the driver of the car does not pay any attention to the speed limit

The shortest time in which the driver can complete the operation is approximately 6.32456 seconds

Explanation:

The given parameters are;

The speed of the truck and the car = 35 mi/h ≈ 51.33 ft./s

Let "t" represent the time it takes the car to pass the truck by 40 ft., we have;

The distance covered by the truck = 35 mi/h × t

The distance covered by the car = 35 mi/h × t + 80 ft = 51.33

Let the distance over which the car accelerates = d

We have;

d = 51.33×t₁ + 1/2×5×t₁²

51.33·t + 80 - d = (51.33 + 5·t₁)·(t - t₁) - 1/2·20·(t - t₁)²

51.33 = 51.33 + 5·t₁ - 20·(t - t₁)

∴ 5·t₁ = 20·(t - t₁)

5·t₁ = 20·t - 20·t₁

25·t₁ = 20·t

t₁ = 4·t/5

(51.33 + 5·t₁)·(t - t₁) - 1/2·20·(t - t₁)² + 51.33×t₁ + 1/2×5×t₁² = 51.33·t + 80

We get;

(51.33 + 5·(4·t/5))·(t - (4·t/5)) - 1/2·20·(t - (4·t/5))² + 51.33×(4·t/5) + 1/2×5×(4·t/5)² = 51.33·t + 80

(51.33 + 4·t)·(t/5) - 2·t²/5 + 51.33 × (4·t/5) + 8·t²/5 = 51.33·t + 80

51.33×t/5 + 51.33×4·t/5 + 4·t²/5 - 2·t²/5 + 8·t²/5 = 51.33·t + 80

51.33·t + 10·t²/5 = 51.33·t + 80

2·t² = 51.33·t + 80 - 51.33·t = 80

t² = 80/2 = 40

t = √40 = 2·√10 ≈ 6.32456

The shortest time in which the driver can complete the operation is "t" ≈ 6.32456 seconds

conduction, convection, radiation

the maxium height is 34m/s