Answer:
The shortest time in which the driver can complete the operation is approximately 6.32456 seconds
Step-by-step 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