Question

and
А train staring from rest,
accelerates uniformly
attains a velocity 48 kmh in
2 minutes. It travels at this
speed for 5 minutes. Finally, it
moves with uniform retardation
and is stopped after 3 minutes
Find the total distance travelled
by the train.

Answer 1

Given:-

First Case

Initial Velocity of train = 0m/s

Final Velocity of train = 48km/h→ 13.33m/s

Time taken = 5min = 300s

Second Case:-

Initial Velocity = 13.33m/s

Final Velocity = 0m/s

Time = 3min → 180s

To Find:-

Total Distance travelled by train

Formulae used:-

v = u + at

v² - u² = 2as

Where,

v = Final Velocity

u = Initial Velocity

a = Acceleration

t = Time

s = Distance

Now,

First Case:-

→ v = u + at

→ 13.33 = (0) + a × 180

→ 13.33 = 180a

→ a = 13.33/180

→ a = 0.074m/s²

Therefore,

Distance travelled by train in 5minute

→ v² - u² = 2as

→ (13.33)² - (0) = 2 × 0.074 × s

→ 177.68 = 0.88s

→ s = 177.68/0.88

→ s¹ = 201.9m

Second Case:-

→ v = u + at

→ 0 = 13.33 + a × 180

→ -13.33 = 180a

→ a = -13.33/180

→ a = -0.0740

Therefore,

→ v² - u² = 2as

→ (0)² - (13.33)² = 2 × -0.074 × s

→ -177.68 = -0.1481 × s

→ -177.68 = - 0.148s

→ s = -177.68/;0.148

→ s² = 1200.55

Hence, The Distance travelled is s¹ + s² → 201.9 + 1200.55 → 1402.4m.