Question

15. 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-2 2
(iii) 4600 m, (iv) 17.69 m s-1

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Answer 1

Explanation:

1) maximum velocity

v max = u + at

= 0 + 2(10)

= 20 m/s

2)retardation

v = u + at (here v=0 and u=20)

0 = 20 + a (50)

-20 = 50a

a = -20/50

= 0.4m/s^2

3)total distance travelled

v^2 - u^2 = 2as

4) average velocity =

total distance travelled / total time taken

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