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## Official (Closed) – Non Sensitive MEF Tutorial 2 Q3 A train with a maximum speed of 29.17 m/s has an acceleration rate of 0.2

Question

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.

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Physics
2 months
2021-09-05T07:41:38+00:00
2021-09-05T07:41:38+00:00 1 Answers
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## Answers ( )

Answer: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 + at₁t₁ = (Vf – Vi)/at₁ = (29.17 m/s – 0 m/s)/(0.25 m/s²)t₁ = 116.68 sUsing 2nd equation of motion:

s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²s₁ = 1701.78 m = 1.7 kmNow, 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 + at₂t₂ = (Vf – Vi)/at₂ = (0 m/s – 29.17 m/s)/(- 0.7 m/s²)t₂ = 41.67 sUsing 2nd equation of motion:

s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²s₂ = 607.78 m = 0.6 kmSince, 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 = vt₃t₃ = s₃/vt₃ = (4700 m)/(29.17 m/s)t₃ = 161.12 sSo, the minimum running time will be:

t = t₁ + t₂ + t₃t = 116.68 s + 41.67 s + 161.12 st = 319.47 s