Question

A train starts from rest and accelerates uniformly at a rate of 2 m s^-2 for 10 s. It then maintains a constant speed for 200 s. The brakes are then applied and the train is uniformly retarded and comes to rest in 50 s. Find

(i) The maximum velocity reached, 

(ii) The retardation in the last 50 s, 

(iii) The total distance travelled,

(iv) The average velocity of the train.

Answer 1

(i) For the first 10 s, initial velocity u = 0

Acceleration a = 2 m/s²

Time taken t = 10 s

Let 'v' be the maximum velocity reached.

Using the first equation of motion

v = u + at

We get

V = (0) + (2) (10) = 20 m/s^-1

(ii) For the last 50 s: Final velocity = 0 m/s, initial velocity = 20 m/s.

Acceleration = (Final velocity - Initial velocity)/time

 = (0 - 20)/50 = -0.4 m/s²

Retardation = 0.4 m/s^-2

(iii) Total distance travelled = Distance travelled in the first 10 s + Distance travelled in 200 s + Distance travelled in last 50 s

 

Distance travelled in first 10s (s₁) = ut + (1/2) at²

S₁= (0) + (1/2) (2) (10)²

S₁= 100 m

 

Distance travelled in 200s (s₂) = speed × time

S₂ = (20) (200) = 4000 m

 

Distance travelled in last 50s (s₃) = ut + (1/2) at²

Here, u = 20 m/s, t = 50 s and a = -0.4 m/s²

S₃= (20)(50) + (1/2) (-0.4) (50)²

S₃= 1000 - 500

S₃= 500 m

 

Therefore, total distance travelled = S₁ + S₂ + S₃ = 100 + 4000 + 500 = 4600 m

 

(iv) Average velocity = Total distance travelled/total time taken

 = (4600/260) m/s

 = 17.69 m/s^-1