Question

In designing subway trains and stations, engineers want to optimize the time and energy use between stations. A train can comfortably accelerate at 5 ft/s2 without causing its passengers to fall over. When coasting, the train will slow down at an acceleration of -2 ft/s2. The train begins at rest and must come to a complete stop at the destination. If two train stations are 2 miles apart, determine the fastest time that the train can arrive at the second station given the provided accelerations. Also, determine the maximum velocity obtained. Draw the a-t, v-t, and s-tgraphs for the train. (122 s, 174 ft/s)

Answer 1

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 .