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, and (iv) the
average velocity of the train.
Ans. (i) 20 m s-1, (ii) 0.4 m s?,
(iii) 4600 m, (iv) 17.69 m s-1

show the procedure​

Answer 1

u = 0

v = u + at

v = 0 + 2(10) = 20 m/s

max Velocity = 20 m/s

Distance covered in 10 Sec

S = ut + (1/2)at² = 0 + (1/2)(2)(10²) = 100 m

Distance covered in next 200 s

S = 20 * 200  = 4000 m  ( acceleration = 0  as constant speed)

0 = 20 + a(50)

=> a = -0.4  m/s

-ve acceleration is retardation so 0.4 m/s is retardation

Distance covered in 50 Sec

S = 20*50 + (1/2)(-0.4)(50)²

S = 1000 - 500

S = 500 m

Total Distance Covered 100 + 4000 + 500 = 4600 m

Total time = 10 + 200 + 50 = 260 sec

Average speed = 4600/260 = 17.7 m/s