Answer:
Explanation:
Let the time required for acceleration a₁ and deceleration a₂ be t₁ and t₂ .
Since final velocity during acceleration and initial velocity during deceleration are same
a₁ t₁ = a₂ t₂
5t₁ = 2 t₂ ------------------------------------------ ( 1 )
Distance travelled during acceleration = 1/2 a₁t₁²
= 1/2 x 5 x t₁² = 2.5 t₁²
Distance travelled during deceleration = 1/2 a₂t₂²
= 1/2 x 2 x t₂² = t₂²
Total distance travelled = 2 miles = 2 x 1760 x 3 ft = 10560
2.5 t₁² + t₂² = 10560
2.5 ( 2t₂ / 5 )² + t₂² = 10560
.4 t₂² + t₂² = 10560
1.4 t₂² = 10560
t₂ = 86.85 s
t₁ = 2t₂ / 5 = 34.75 s
t₁ + t₂ = 121.6 = 122 s
Total time taken = 122 s .
maximum velocity = a₁t₁
= 5 x 34.75 = 173.75 = 174 m/s .