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.​

Answer 1

here u go

Train accelerate from start for 10 s

so velocity reached at 10s is a×t= 2×10=20m/s

1) 20m/s is max velocity as after this the train runs on constant speed

2) the train is retarded for 50s from this speed to rest so retardation is velocity /time=20/50=

-0.4m/s2

3) distance travelled is total of distance travelled with accelaration+distance with constant speed+ distance with retardation

distance with acceleration is ut+1/2at2

=1/2*2*10*10=100m

distance with constant speed is s×t=20×200=4000m

distance with retardation is ut+1/2at2

20×50-1/2×.4×50×50

=1000-500

=500m

total distance =100+4000+500=4600m

4) average velocity= total distance/total time

= 4600/260

=17.69m/s