A train starts from rest and accelerates uniformly at A rate of 2ms-1 for 10-s.It then maintains A constant speed for 200 s. Tge 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, and (iv) the average velocity of the train. 

Starting from rest , if accelerated for 10 s with acceleration 2 m/s² ,

then velocity after 10 s is determined from the following equation

 

v = a × t  = 2  ×  10  = 20 m/s

 

Distance travelled in 10 s ,  S₁ = (1/2) a t² = 0.5 × 2 × 10 × 10 = 100 m

 

Distance travelled in 200 s,  S₂ =  velocity × time = 20 × 200 = 4000 m

 

Let a_r be the retardation to stop the train in 50 s , then retardation is detrimined from

 

v - ( a_r × t ) = 0    or   a_r = v / t  = 20 /50  = 0.4 m/s²

 

Distance travelled in 50 s ,  S₃ = ( v × t ) - ( 1/2) × a_r × t² 

 

S₃ = ( 20 × 50 ) - ( 0.5 × 0.4 × 50 × 50 ) = 500 m

 

Answers :-

 

(i) Maximum velocity = 20 m/s

 

(ii) Retardation in last 50 s  = 0.4 m/s²

 

(iii) Total distance = S₁ + S₂ + S₃ = 100 + 4000 + 500 = 4600 m

 

(iv) average velocity = net Displacement / total time = 4600 / (10+200+50) = 17.69 m/s