Official (Closed) - Non Sensitive

MEF Tutorial 2 Q3

A train with a maximum speed of 29.17 m/s has an

acceleration rate of 0.25 m/s2 and a deceleration

rate of 0.7 m/s2. Determine the minimum running

time, if it starts from rest at one station and stops

at the next station 7 km away.

The minimum running time is 319.47 s.

Explanation:

First we find the distance covered and time taken by the train to reach its maximum speed:

We have:

Initial Speed = Vi = 0 m/s (Since, train is initially at rest)

Final Speed = Vf = 29.17 m/s

Acceleration = a = 0.25 m/s²

Distance Covered to reach maximum speed = s₁

Time taken to reach maximum speed = t₁

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi) / a

t₁ = (29.17 m/s - 0 m/s) / (0.25 m/s²)

t₁ = 116.68 s

Using 2nd equation of motion:

s₁ = (Vi) (t₁) + (0.5) (a) (t₁) ²

s₁ = (0 m/s) (116.68 s) + (0.5) (0.25 m/s²) (116.68 s) ²

s₁ = 1701.78 m = 1.7 km

Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.

We have:

Final Speed = Vf = 0 m/s (Since, train is finally stops)

Initial Speed = Vi = 29.17 m/s (The train must maintain max. speed for min time)

Deceleration = a = - 0.7 m/s²

Distance Covered to stop = s₂

Time taken to stop = t₂

Using 1st equation of motion:

Vf = Vi + at₂

t₂ = (Vf - Vi) / a

t₂ = (0 m/s - 29.17 m/s) / ( - 0.7 m/s²)

t₂ = 41.67 s

Using 2nd equation of motion:

s₂ = (Vi) (t₂) + (0.5) (a) (t₂) ²

s₂ = (29.17 m/s) (41.67 s) + (0.5) ( - 0.7 m/s²) (41.67 s) ²

s₂ = 607.78 m = 0.6 km

Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.

The remaining distance is:

s₃ = 7 km - s₂ - s₁

s₃ = 7 km - 0.6 km - 1.7 km

s₃ = 4.7 km

Now, for uniform speed we use the relation:

s₃ = vt₃

t₃ = s₃/v

t₃ = (4700 m) / (29.17 m/s)

t₃ = 161.12 s

So, the minimum running time will be:

t = t₁ + t₂ + t₃

t = 116.68 s + 41.67 s + 161.12 s

t = 319.47 s